Surprising identities / equations What are some surprising equations/identities that you have seen, which you would not have expected?
This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc.
I'd request to avoid 'standard' / well-known results like $ e^{i \pi} + 1 = 0$.
Please write a single identity (or group of identities) in each answer.
I found this list of Funny identities, in which there is some overlap.
 A: Where $\varphi = \frac{1 + \sqrt{5}}{2}$ a golden ratio, $$\int_0^\infty\frac{1}{(1+x^\varphi)^\varphi}\mathrm dx = 1.$$
This follows immediately from the substitution $t=[x^{\varphi}(1+x^{\varphi})^{-1}]^{\varphi}$.
Proof (below) by filmor

 $$\begin{align} \int_0^\infty\frac{1}{(1+x^\varphi)^\varphi}\mathrm dx &= \varphi^{-1}\int_0^\infty\frac{y^{\varphi^{-1} - 1}}{(1+y)^\varphi}\mathrm dy \\ &= \varphi^{-1}\int_0^\infty\frac{y^{\varphi - 2}}{(1+y)^\varphi}\mathrm dy \\ &= \varphi^{-1} B\bigl(\varphi - 1, 1\bigr) \\ &= \varphi^{-1}\frac{\Gamma(\varphi-1)\ \Gamma(1)}{\Gamma(\varphi)} \\ &= \varphi^{-1} \frac{1}{\varphi - 1} = 1 \end{align}$$

One more thing with golden ratio: by Ramanujan,
$$r=\dfrac{e^{-2\pi/5}}{1 + \dfrac{e^{-2\pi}}{ 1 + \dfrac{e^{-4\pi}}{1 + \cdots}}} =  \sqrt{ \sqrt{5}\varphi} - \varphi$$
and even more bizarrely (found based on the work of Vidunas), the hypergeometric function $N=\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},1\big)$ is a deg-80 algebraic number given by,
$$N=\frac{1}{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{1/20}}$$
A: There are many involving infinite sums of number theoretic functions:
$$\sum_{i = 1}^\infty \frac{\phi(i)}{i^k} = \frac{\zeta(k - 1)}{\zeta(k)}$$
$$\sum_{i = 1}^\infty \frac{\tau(i)}{i^k} = \zeta(k)^2$$
$$\sum_{i = 1}^\infty \frac{\sigma(i)}{i^k} = \zeta(k - 1)\zeta(k)$$
$$\sum_{i = 1}^\infty \frac{\mu(i)}{i^k} = \frac{1}{\zeta(k)}$$
And for some reason, the Riemann zeta function pops up in each. (I have no idea if these are well-known or not, I just thought they were very surprising, because I learned about these functions in a very discrete, number theory context, and the zeta function in, well, not that.)
A: This is the most surprising result
that I am the discoverer of.
Consider the diophantine equation
$$x(x+1)...(x+n-1) -y^n = k$$
where $x, y, n,$ and $k$ are integers,
$x \ge 1$,
$y \ge 1$,
and
$n \ge 3$.
I was led to consider considering this
by trying to generalize the
Erdos-Selfridge result
that the product of consecutive integers
could never be a power.
I phrased this as
"How close and how often
can the product of 
$n$ consecutive integers
be to an $n$-th power?"
Looking at this equation,
it seemed reasonable to think that,
for fixed $k$ and $n$,
there were only a finite number of
$x$ and $y$ that satisfied it.
This was not too hard to prove.
What greatly surprised me was that
I was able to prove that
for any fixed $k$,
there were only a finite number of
$n$, $x$, and $y$ that satisfied it.
The proof went like this:
I first showed that
any solution must have
$y \le |k|$.
This was moderately straightforward,
and involved considering the three cases
$y < x$, $x \le y \le x+n-1$,
and $y \ge x+n$.
The next step really surprised me.
I showed that
$n < e|k|$,
where $e$ is the good old base of natural logarithms.
The proof was amazingly (to me) simple.
Since $y \le |k|$
and
$2(n/e)^n < n!$,
$\begin{align}
2(n/e)^n 
&< n!\\
&\le x(x+1)...(x+n-1)\\
&= y^n+k\\
&\le |k|^n+|k|\\
&\le |k|^n+|k|^n\\
&= 2|k|^n\\
\end{align}
$
so
$n < e |k|$.
I still remember staring at this
in disbelief,
over forty years later.
A: I have always been fascinated by Leibniz's formula for $\pi$:
$$\dfrac{\pi}{4}=\sum\limits_{n=0}^\infty \dfrac{1}{2n+1}\times(-1)^{n}=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}\dots$$
This can be used to determine the exact value of $\pi$, which is what makes it interesting.
$$\displaystyle \boxed{\pi=4\sum\limits_{n=0}^\infty \dfrac{1}{2n+1}\times(-1)^{n}=4-\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{7}+\dfrac{4}{9}-\dfrac{4}{11}\dots}$$
A: The Fibonacci sequence occurs in the decimal expansion of some "special" fractions.
$\begin{array}{ccccccccc}
\frac{1}{89} &= &0.\color{blue}{0} \\
             &+ &   &\color{blue}{1} \\
             &+ &   &  &\color{blue}{1} \\
             &+ &   &  &  &\color{blue}{2} \\
             &+ &   &  &  &  &\color{blue}{3} \\
             &+ &   &  &  &  &  &\color{blue}{5} \\
             &+ &   &  &  &  &  &  &\color{blue}{8} \\
             &+ &   &  &  &  &  &  &  &\color{blue}{13} \\
             &+ &   &  &  &  &  &  &  &  &\ddots \\
\end{array}\\
\begin{array}{cc}\frac{1}{89} &=0.\color{blue}{011235}\color{red}{9}\ldots
\end{array}$
Eventually the pattern is destroyed by carries in the decimal digits of the sum. If you want to go further, simply take a different fraction:
$\begin{array}{lllllllll}
\frac{1}{9899} &= &0.
&\underbrace{00}_{F_0}
&\underbrace{01}_{F_1}
&\underbrace{01}_{F_2} 
&\underbrace{02}_{F_3}
&\underbrace{03}_{F_4}
&\underbrace{05}_{F_5} 
&\underbrace{08}_{F_6}
&\underbrace{13}_{F_7} 
&\underbrace{21}_{F_8}
&\underbrace{34}_{F_9} 
&\underbrace{55}_{F_{10}} 
&\ldots
\\
\end{array}$
Again the pattern from this point forward is obscured by carries in the sum (the next two digits are $90$ not $89$).
These patterns are a consequence of the fact that the generating function for the Fibonacci sequence (with $F_0=0, F_1=1$) is
$$f(x)=\frac{x}{1-x-x^2}$$
and taking $x=0.1,0.01,0.001,\ldots$ ensures that the terms in the power series expansion are synonymous with places in the decimal expansion. These values of $x$ are identified with the "special" denominators $89,9899,998999,\ldots$ which are really just entries in the sequence $a_n=10^{2n}-10^n-1$ for $n=1,2,3,\ldots$.
A: I was really amazed by discovering that the squared arcsine function has a pretty nice Taylor series at the origin: $$\arcsin^2(z)=\frac{1}{2}\sum_{n\geq 0}\frac{(2z)^{2n}}{n^2\binom{2n}{n}}$$
Even more amazed by the variety of techniques one may employ to prove such identity: combinatorial convolutions, hypergeometric transformations, (poly)logarithmic integrals, the Lagrange inversion theorem, the residue theorem, Legendre polynomials, Euler's Beta function, creative telescoping... They're all pretty interesting.
A: If $A+B+C=180^\circ$ then
$$\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$$
A: Here is a mathematical scherzo.
$$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3.$$
A: $\displaystyle\sum_{k=1}^{24} k^2=70^2$ is novel.
A: $$
{a \over b} = {c \over d}
\quad\Longrightarrow\quad 
{a + b\over a - b} = {c + d \over c - d}
$$
A: $$\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^\pi}dx=\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^e}dx$$
A: $$\sum_{k=-\infty}^\infty 2^k = 0$$ as can be shown from the regularization $\sum\limits_{k=1}^\infty 2^k=-1$. I'm wondering whether this is not actually the case for all ("sensible") two-sided regularizations, see here

Too explain this sum, note how for $|q|<1$ the geometric series $1+q+q^2+...$ converges to $\frac1{1-q}$. This works fine for the negative powers of two, i.e. $q=\frac12$ such that $\sum\limits_{k=-\infty}^02^k=\sum\limits_{k=0}^\infty\left(\frac12\right)^k=2$. Regularization now basically consists of stating "Ok, outside the convergence region (the positive powers of two in this case) just claim $1+q+q^2+...$ is still "equal" to $\frac1{1-q}$", i.e. $\sum\limits_{k=0}^\infty 2^k "=" \frac1{1-2}=-1$. Subtract the $1=2^0$ counted twice from these two sums to get above result.
We Theoretical Physicists tend to do things like this regularly, which mostly means we were too eager on swapping $\lim$s at some point before ;)
A: $$\sum_{i=0}^N {{N}\choose{i}}=2^N$$
A: $$
2^{67}-1 = 193,707,721 × 761,838,257,287
$$
This identity was found by Cole in the early 20th century.  He later said that the result had taken him "three years of Sundays" to find. 
There's also the fact that:
 $2^{127} -1 $ is indeed prime, as Mersenne claimed. This was the largest known prime number for 75 years, and the largest ever calculated by hand. Édouard Lucas proved its primality in 1876.
A: The infinite Fibonacci sequence $F = (0, 1, 1, 2, 3, 5, 8, \dots)$ and the infinite Fibonacci string $S = 1011010110110 \dots$ are related by the following remarkable identity for all real or complex $\beta$ such that $\ |\beta| > 1$:  
$$[0 \ ; \ \beta^{F_0}, \beta^{F_1}, \beta^{F_2}, \dots] = (\beta - 1) \cdot (0.S)_\beta$$
where $[0 \ ; \ \beta^{F_0}, \beta^{F_1}, \beta^{F_2}, \dots]$ denotes the continued fraction
$$\frac{1}{\beta^{F_0} + \frac{1}{\beta^{F_1} + \frac{1}{\beta^{F_2} +  \cdots}}} $$
and $(0.S)_\beta = (0.1011010110110 \dots)_\beta$ denotes the number obtained by reading $0.S$ as a "base-$\beta$ numeral"; that is, $(0.S)_\beta$ denotes the sum of the infinite series 
$$S[1] \beta^{-1} + S[2]\beta^{-2} + S[3]\beta^{-3} + \cdots$$ 
where $S[n]$ is the $n$th element of string $S$.
E.g., the so-called rabbit constant $(0.S)_2$ = 0.709803... in decimal , $(0.S)_\pi$ = 0.362011... in decimal, etc. 
($F$ and $S$ are also related by the fact that both are generated by recursions of the form $x_{n+1} = x_{n} + x_{n-1} ;\ x_0 = 0, \ x_1 = 1$, where the $+$ is interpreted in one case as arithmetic addition, and in the other case as string concatenation.)
A: When I began my serious encounter with number theory and looked at properties of prominent combinatorical matrices I found this identity. This impressed me so much (even a bit philosophically) that I wanted to printed it on a t-shirt (but the white-on-black printing was then too expensive). The german phrase means "the exponential of the counting is the binomial" 
Here is, how it looked asymptotically:

A: I know it's incredibly simple, but I'm always awed by
$$
2+2 = 2 \cdot 2 = 2^2 = \;^2 2.
$$
Two is where addition, multiplication and exponentiation meet. And: tetration.
A: Given a polynomial $p(x)$ of degree $n$, let $a$ be the leading coefficient.  Then:
$$\sum_{k=0}^n (-1)^k{n\choose k}p(x-k)=an!$$
This happens to be equivalent to:
$$p^{(n)}(x)=an!$$
where $p^{(k)}$ is the $k$th derivative of $p(x)$.
The surprising part is that the sum can actually obtain the leading coefficient without any remaining reference to the polynomial aside from the factorial of the degree.
A: $$\sum_{k=1}^{\infty}k^{-k}=\int_0^1x^{-x}\mbox{ d}x=\mathrm{Sophomore's}\mbox{ }  \mathrm{dream}$$ 
A: $y(x)=\left \{ \frac{x- \frac {\left \lceil \frac {\sqrt {1+8x}-1} {2} \right \rceil \left ( 1-\left \lceil \frac {\sqrt {1+8x}-1} {2} \right \rceil \mod2 \right)} {2}}  {\left \lceil \frac {\sqrt {1+8x}-1} {2} \right \rceil}\right \}$
This function has the following slopes: 1 at [0,1), 2 at [1,3) and so on
A: Let $p_n$ be the probability that a random permutation in the symmetric group $S_n$ doesn't have fixed points. Then $\lim_{n\to\infty}p_n=\frac{1}{e}$. 
I was amazed the first time I saw this exercise!
A: It's funny noone mentioned the hockey-stick identity, partial sum of columns in a Pascal triangle:

$$
\sum_{k=0}^{m}\binom{n+k}{k}=\binom{n+m+1}{n}
$$
http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity
A: $$
\sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1-\frac12-\frac13-\frac15+\frac16-\frac17+\frac1{10}-\frac1{11}-\frac1{13}+\frac1{14}+\frac1{15}-\cdots=0
$$
This relation was discovered by Euler in 1748 (before Riemann's studies on the $\zeta$ function as a complex variable function, from which this relation becomes much more easier!).
Another notable relation is the following, on the partition function, due to Ramanujan:
$$
p(n)=\frac1{\pi\sqrt2}\sum_{k=1}^{N}\sqrt k\left(\sum_{h\mod k}\omega_{h,k}e^{-2\pi i\frac{hn}{k}}\right)\frac d{dn}\left(\frac{\cosh\left(\frac{\pi\sqrt{n-\frac1{24}}}{k}\sqrt{\frac23}\right)-1}{\sqrt{n-\frac1{24}}}\right)+O\left(n^{-\frac14}\right)\;.
$$
Then one of the most impressive formulas is the functional equation for the $\zeta$ function, in its asimmetric form: it highlights a very very deep and smart connection between the $\Gamma$ and the $\zeta$:
$$
\pi^{-\frac s2}\Gamma\left(\frac s2\right)\zeta(s)=
\pi^{-\frac{1-s}2}\Gamma\left(\frac{1-s}2\right)\zeta(1-s)\;\;\;\forall s\in\mathbb C\;.
$$
A: Plot the graphs of the functions $$f(x)=\dfrac{2(x^2+|x|-6)}{3(x^2+|x|+2)}+\sqrt{16-x^2}$$ and $$g(x)=\dfrac{2(x^2+|x|-6)}{3(x^2+|x|+2)}-\sqrt{16-x^2}$$ in $x\in[-4,4]$ on the same plane. 

A: This one really surprised me:
$$\int_0^{\pi/2}\frac{dx}{1+\tan^n(x)}=\frac{\pi}{4}$$
A: This bit of notational juggling may cause one do double take...

$$\huge \sqrt[\sqrt{2}]{2} = \sqrt{2}^\sqrt{2}$$

By definition, the LHS is the number $x$ such that $x^\sqrt{2} = 2$.  It is simple to check that the RHS also has this property. 
A: Something I recently saw on Abstruse Goose (although I don't recall the exact link).
$$10^2+11^2+12^2=13^2+14^2$$
Moreover, one can easily prove that this is the only sequence of five consecutive positive numbers which have this property!
A: The continued fraction of The Golden Ratio:
$\frac{1+\sqrt5}{2}=[1;,1,1,1,\dots]=[1,\bar1]$
Also:
$\frac{1+\sqrt5}{2}=\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}$.
A: $$
{\large\sqrt{\vphantom{\Large A}\,\color{#ff0000}{20}\color{#0000ff}{25}\,}\, = 45 = \color{#ff0000}{20} + \color{#0000ff}{25}}
$$
A: $$\sin \pi x=\pi x\prod_{n=1}^{\infty}\left(1-\frac {x^2}{n^2}\right).\quad \text {(L. Euler).}$$ Obviously the LHS and RHS have the same set of zeroes but that alone does not imply equality.  And putting $x=1/2$ into it, we derive the Wallis product for $\pi, $ which itself is remarkable, especially as Wallis was Newton's immediate predecessor in the "Lucas chair" and obtained his product without the full generality of  the methods of calculus developed by Newton..
A: $$\ \ \ \ \ 2592=2^59^2\ \ \ \ \ $$
A: Easy geometric series but I found this one charming when I found out:
1/7 = 0,142857...
    = 0,14 +
      0,0028 +
      0,000056 +
      0,00000112 +
      0,0000000224 + ... (double the value and shift it by two spaces)

A: Do logic answers count?  I like the Drinker Paradox, which isn't really a paradox but actually a theorem of logic:
$\exists x.\ [D(x) \rightarrow \forall y.\ D(y)]$
For every bar there is a person for whom, if that person is drinking, then everyone is drinking.
A: $$\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}dx=\dfrac{\pi e}{4!}$$
A: $ \tan 10^\circ  = \tan 20^\circ \times \tan 30^\circ \times \tan 40^\circ $.
$\tan 80^\circ = \tan 70^\circ \times \tan 60^\circ \times \tan 50^\circ $.   
A: One more with continued fractions. In 2003 there was a discussion in sci.math about the continued fractions of powers of $e$ - if I recall correctly, then that of even powers are somehow folklore. 
But examining the pattern to the depth we came to the following infinite continued fraction with a variable parameter:
$$ \operatorname{cfe}(x)= [1,\tfrac1x-1,1, \quad 1,\tfrac3x-1,1, \quad 1,\tfrac5x-1,1, \quad \ldots ]$$
where the pattern is easily recognizable.
Then "generalize" the continued fraction and allow irrational values for $x$. Then
$$ x = \operatorname{cfe}( \ln(x) )  \qquad \qquad x \ne 1$$ 
or 
$$ x= 1+\cfrac{1}
{(\tfrac1{\ln x}-1) + \cfrac{1}
  {1+\cfrac{1}
   {1+\cfrac{1}
     {(\tfrac3{\ln x}-1) + \cfrac{1}
       {1+\cfrac{1} {1+\cfrac{1}
{(\tfrac5{\ln x}-1) + \cfrac{1}
  {1+\cfrac{1}
   {...}}}}
}}}}}$$ 
A: $$\int_0^{\frac{\pi}{2}} x\ln(\tan(x))\ dx=\frac{7}{8}\zeta(3)$$
A: I'm a fan of approximations, and I ran into this one the other day:
$$
\Gamma^{(k)}(1) \sim (-1)^k\, \Gamma(k+1) \quad \text{as } k \to \infty.
$$
The form is interesting in that it relates the $k^\text{th}$ derivative of the function at $1$ to the value of the function at $k+1$.
The approximation isn't too bad either; the relative error is on the order of $2^{-k}$, i.e.
$$
\Gamma^{(k)}(1) = (-1)^k\, \Gamma(k+1) \left[1 + O\!\left(2^{-k}\right)\right] \quad \text{as } k \to \infty.
$$
A: I apologize if this is already here.  I thought it was but I can't find it, so I must have seen it somewhere else on the site.
$$\begin{matrix}
f(x) & \displaystyle\int f(x)dx \\[6pt] \hline
x^2 & \dfrac{x^3}{3} \\[6pt]
x & \dfrac{x^2}{2} \\[6pt]
1 & x \\[6pt]
\dfrac{1}{x} & \color{red}{\log(x)} \\[6pt]
\dfrac{1}{x^2} & -\dfrac{1}{x} \\[6pt]
\dfrac{1}{x^3} & -\dfrac{1}{2x^2} \\[6pt]
\dfrac{1}{x^4} & -\dfrac{1}{3x^3}
\end{matrix}$$
$\displaystyle{\int x^n dx} = \frac{x^{n+1}}{n+1}$
$\displaystyle{\lim_{n \rightarrow -1} \left(\frac{x^{n+1}}{n+1}\right)} = \log(x)$?
No. Let $g(x,n)=\frac{x^{n+1}}{n+1}$.
For $x\in(0,\infty)$, you have:
$g(x,n)>0$ as $n\rightarrow -1$ from above, and 
$g(x,n)<0$ as $n\rightarrow -1$ from below, but 
$\log(x)<0$ on $x\in(0,1)$ and $\log(x)>0$ on $x\in(1,\infty)$.
I wish I understood this better.
A: $$\frac{11}{10}\cdot\frac{1111}{1110}\cdot\frac{111111}{111110}\cdot\frac{11111111}{11111110}\cdots
=1.101001000100001000001\cdots$$
A: I got this integral,
$$\int_{-\infty}^{+\infty}\frac{\mathrm dt}{(\phi^n t)^2+\pi^2(F_{2n+1}-\phi F_{2n})(e^{\gamma}t^2+t-1)^2}=1$$
Where $F_{n}$ is the Fibonacci number,$\phi$ is the Golden ratio and $\gamma$ is the Euler's constant.
A: Let me add my personal favorite one here:
$$\sum_{\substack{m, n \geq 1\\\gcd(m,n)=1}} \frac{1}{m^2n^2}=\frac{5}{2}.$$
You can generalize this into
$$\mathop{\sum_{n_1=1}^\infty\cdots\sum_{n_j=1}^\infty}\limits_{(n_1,\dots,n_j)=1} \frac1{n_1^{k_1} \cdots n_j^{k_j}}=\frac{\zeta(k_1)\zeta(k_2)\cdots \zeta(k_n) }{\zeta(k_1+k_2+\cdots+k_n)},$$
where $\zeta$ is the Riemann's zeta function.
A: If $x^{x} = x+1$, then
$x^{x^2+x} = (x^2+x)^{x}$
A: $$
\int_0^1 \frac{\ln(1+t^{4+\sqrt{15}})}{1+t}dt= -\frac{\pi^2}{12}(\sqrt{15}-2)+\ln 2\cdot \ln(\sqrt{3}+\sqrt{5})+\ln\frac{1+\sqrt{5}}{2}\cdot \ln(2+\sqrt{3}) 
$$
For references, see http://ega-math.narod.ru/Chowla/index.htm (there is a scan of a paper of Herglotz where it is proved).
A: The square root of 2 is also the only real number other than 1 whose infinite tetrate is equal to its square...
$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}=2.$$
A: I find this identity due to Euler particularly striking (and not obvious at all):
$$\prod_{n=1}^\infty (1-x^n) = \sum_{k=-\infty}^\infty (-1)^k\,x^{p(k)}$$
where the $p(k) = \dfrac{k(3k-1)}{2}$ are the generalized pentagonal numbers. This is what these numbers look like us  for $1 \leq k \leq 5$,

[image created by Aldoaldoz]
A: Tetration :
consider the tower of taking infinite powers :
           $x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}}$ . 
At first its seems big mystery and undefined one for lots of real numbers.
Surprising fact is its indeed converges in an closed interval which is bounded by the fancy real numbers $e^{-e}$, $e^\frac{1}{e}$ 
So       $x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}}$ converges for $ x \in [e^{-e}, e^\frac{1}{e} ] $
A: If $a,b,c,d$ are in arithmetical progression, then
$$\frac{d^2-a^2}{c^2-b^2}=3.$$
A: $3^3 + 4^4 + 3^3 + 5^5 = 3435$
$1^1=1$ is the only other such number.
A: $$ \begin{align}\frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}   \\\,\\\,\\
 \frac{\pi}{4} = 5 \arctan \frac{1}{7} + 2 \arctan \frac{3}{79}\end{align}$$
Both can be shown easily using polar form, complex multiplication. 
A: This formula thrills me and stirs my mind as nothing could for years...
$$\int^\infty_{0}\!\!e^{-3\pi x^2}\frac{\sinh(\pi x)}{\sinh(3 \pi x)}\,dx=\frac{1}{e^{2\pi/3}\cdot \sqrt3} \sum^\infty_{n=0}\frac{e^{-2n(n+1)\pi}}{(1+e^{-\pi})^{2}(1+e^{-3\pi})^{2}\cdots(1+e^{-(2n+1)\pi})^{2}}$$
A: I find the Young-Frobenius identity, found on p. 8 here, surprising:
A partition $\lambda\vdash n $ of an integer $n\geq0$, i.e. a sequence $(\lambda_{1}, \cdots,\lambda_{k})$ with $\lambda_{1}\geq \cdots \geq \lambda_{k}>0$ and $\lambda_{1} + \cdots + \lambda_{k}=n$, can be identified with a diagram consisting of $k$ left-justified rows of boxes, where row $i$ (starting from the top) has $\lambda_{i}$ boxes, called a Young diagram of size $n$. For some Young diagram $\lambda$ of size $n$, a Young tableau of shape $\lambda$ is an assignment of integers $1$ through $n$ to the boxes of $\lambda$; a Young tableau is standard if these numbers increase in each row and column. For example, one size-$10$ standard Young tableau of shape $(5,4,1)$ is:

Now let $f^{\lambda}$ denote the number of standard Young tableaux of shape $\lambda$. Then the numbers of standard Young tableaux of each shape of size $n$ satisfy this identity:
$$\sum_{\lambda\vdash n}^{}(f^{\lambda})^{2}=n!$$
A: One of the most beautiful formulas in combinatorics: Cayley's Formula:Number of labelled trees on $n$ vertices $=n^{n-2}$ And here are some interersting, yet not-so-popular combinatorial identities:
Notations:
$F_n = n^{th}$ Fibonacci number $H_n =$ $n^{th}$ Harmonic number; $H_n = 1+ \frac{1}{2}+\frac{1}{3} +...+\frac{1}{n}$ 


*

*$$ \sum_{n \ge 1} {\frac{F_n}{2^n}} =2 $$

*$$ \sum_{n \ge 1} {n \frac{F_n}{2^n}} =10 $$

*$$ \sum_{n \ge 1} {n^2\frac{F_n}{2^n}} =94 $$

*For $0 \le m \le n$,  $$\sum_{k=m}^{n-1} {\binom{k}{m} \frac {1}{n-k}}= \binom{n}{m}(H_n -H_m)$$

A: The most interesting identity I have come across so far is the Mellin Transform of Gauss' Hypergeometric Function with a negative $x$-argument which can be expressed as a combination of Beta and Gamma Functions
$$\mathcal{M}[_2F_1(a,b;c;-x)](s)=B(s,a-s)\frac{\Gamma(b-s)\Gamma(c)}{\Gamma(b)\Gamma(c-s)}$$
which again points out the extremely close relation between all these special functions in general.
A: I believe McShane's identity is very surprising, and is pretty different from the rest of these answers.
Let $\Sigma_{1,1}$ be a once punctured torus with a complete, finite volume, hyperbolic metric $g$. Let $C$ be the set of closed geodesic curves which do not intersect themselves (simple closed curves). Then we have the following identity:
$$ \sum_{c \in C} \frac{1}{1+e^{\mathrm{length}(c)}}=\frac{1}{2}. $$

This is extremely interesting because there are many such metrics up to isometry. In fact the space of such metrics is naturally isomorphic to the modular curve $\mathbb H^2/ \mathrm{PSL}(2,\mathbb{Z})$. In general the set of lengths of simple closed curves will be very different for different metrics. For example, for any $\epsilon >0$ you can find a metric with a $c \in C$ of that length. When $\epsilon$ is small the corresponding term is very close to $1/2$. There are also metrics where every curve in $C$ has length greater than 1.
This identity, and generalizations of this, end up being related to important work by Maryam Mirzakhani, who won the Fields medal.
A: This expression is from Francois Viète: $$\pi = \frac{2}{1} \times \frac{2}{\sqrt2} \times \frac{2}{\sqrt{2+\sqrt2}}\times
 \frac {2}{\sqrt{2+\sqrt{2+\sqrt2}}} \times \cdots$$
A: Some zeta-identies have been much surprising to me.       
Let's denote the value $\zeta(s)-1$ as $\zeta_1(s)$ then
$$ \small \begin{array} {}
1 \zeta_1(2) &+&1 \zeta_1(3)&+&1 \zeta_1(4)&+&1 \zeta_1(5)&+&  ... &=&1\\
1 \zeta_1(2) &+&2 \zeta_1(3)&+&3 \zeta_1(4)&+&4 \zeta_1(5)&+&  ... &=&\zeta(2)\\
             & &1 \zeta_1(3)&+&3 \zeta_1(4)&+&6 \zeta_1(5)&+&  ... &=&\zeta(3)\\ 
             & &            & &1 \zeta_1(4)&+&4 \zeta_1(5)&+&  ... &=&\zeta(4)\\ 
             & &            & &            & &1 \zeta_1(5)&+&  ... &=&\zeta(5)\\ 
 ...         & & & & & & & &... &= &  ...
\end{array}
$$
There are very similar stunning alternating-series relations:
$$ \small \begin{array} {}
1 \zeta_1(2) &-&1 \zeta_1(3)&+&1 \zeta_1(4)&-&1 \zeta_1(5)&+&  ... &=&1/2\\
             & &2 \zeta_1(3)&-&3 \zeta_1(4)&+&4 \zeta_1(5)&-&  ... &=&1/4\\
             & &            & &3 \zeta_1(4)&-&6 \zeta_1(5)&+&  ... &=&1/8\\ 
             & &            & &            & &4 \zeta_1(5)&-&  ... &=&1/16\\ 
 ...         & & & & & & & &... &= &  ...
\end{array}
$$
A: $$\int_0^1\frac{x^4(1-x)^4}{1+x^2}\mathrm{d}x=\frac{22}{7}-\pi$$
It's interesting how something so bizarre on the left hand side yields the tiniest of errors in one of the most famous approximations of $\pi$.
A: This one by Ramanujan gives me the goosebumps:
$$
\frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}.
$$

P.S. Just to make this more intriguing, define the fundamental unit $U_{29} = \frac{5+\sqrt{29}}{2}$ and fundamental solutions to Pell equations,
$$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$
$$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$
$$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$
then we can see those integers all over the formula as,
$$\frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k} =  \frac{1}{\pi} $$
Nice, eh?
A: ${1\over 2} < \left\lfloor \mathrm{mod}\left(\left\lfloor {y \over 17} \right\rfloor 2^{-17 \lfloor x \rfloor - \mathrm{mod}(\lfloor y\rfloor, 17)},2\right)\right\rfloor$
The above is the most interesting inequality in mathematics.  If you plot it so that areas satisfying the inequality are shaded, this is what you get:

This is known as Tupper's self referential formula.
A: This one is one of my favorite:
$$ \log(1+2+3) = \log(1)+\log(2)+\log(3) $$
A: Another one, which occured to me when I began to learn about double-sums in the context of divergent summation. I really had to chew on this, that the sum of the vertical sums can be different from the sum of the horizontal sums... And just different by the exact value of 1. So this had some appeal as another example of Where is the missing 1 in the equation? (From an older essay of mine):     

A: Ramanujan stated this radical in his lost notebook:
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\dots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$
I still don't have any idea on this one.
A: I think this is simple but I want to post it:
$$1 \times 9=9\implies(0+9=9)$$
$$2\times 9=18\implies(1+8=9)$$
$$3\times 9=27\implies(2+7=9)$$
$$4\times 9=36\implies(3+6=9)$$
$$5\times 9=45\implies(4+5=9)$$
$$6\times 9=54\implies(5+4=9)$$
$$7\times 9=63\implies(6+3=9)$$
$$8\times 9=72\implies(7+2=9)$$
$$9\times 9=81\implies(8+1=9)$$
$$10\times 9=90\implies(9+0=9)$$
and also no. are in a pattern $09,18,27,36,45\;$then reverse the no.$54,63,72,81,90$
A: A surprising family of series for $e$ can be derived by algebraically combining the terms in Newton's series expansion:
\begin{equation}
e=\sum_{k=0}^{\infty } \dfrac{1}{k!}=\dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\dfrac{1}{5!}+\ldots.
\end{equation}
Here are a few examples:
\begin{equation}
e=\sum _{k=0}^{\infty } \frac{2k+1}{(2k)!}=\frac{1}{0!}+\frac{3}{2!}+\frac{5}{4!}+\frac{7}{6!}+\frac{9}{8!}+\frac{11}{10!}+\ldots
\end{equation}
\begin{equation}
2e=\sum _{k=0}^{\infty } \frac{k+1}{k!}=\frac{1}{0!}+\frac{2}{1!}+\frac{3}{2!}+\frac{4}{3!}+\frac{5}{4!}+\frac{6}{5!}+\ldots
\end{equation}
\begin{equation}
1/e=\sum _{k=0}^{\infty } \frac{1-2k}{(2k)!}=\frac{1}{0!}-\frac{1}{2!}-\frac{3}{4!}-\frac{5}{6!}-\frac{7}{8!}-\frac{9}{10!}-\ldots~.
\end{equation}
Beyond being pretty, these series converge substantially faster than Newton's series. For more formulas and details on derivation see: http://www.brotherstechnology.com/math/cmj-supp.html
A: Möbius inversion formula may be an example. Also Ramanujan's Partition Congruences were surprising to me when I first saw them.
A: It has surprised me that
$$F_n^\star=\frac{\phi^n-(-\phi)^{-n}}{\sqrt5}$$
where $\phi$ is the golden ratio and $F_n^\star$ is the nth Fibonacci number.  Then, one stumbles upon characteristic equations:
$$a_kF_{n+k}=a_{k-1}F_{n+k-1}+a_{k-2}F_{n+k-2}+\dots+a_0F_n$$
$$\implies F_n=b_kr_k^n+b_{k-1}r_{k-1}^n+\dots+b_1r_1^n$$
where $r$ is a solution of the equation
$$a_kr^k=a_{k-1}r^{k-1}+a_{k-2}r^{k-2}+\dots+a_0$$
assuming the roots do not repeat.  The coefficients are determined based on initial values.
It then surprises me further that this extends to differential equations:
$$a_ky^{(k)}+a_{k-1}y^{(k-1)}+\dots+a_0y=0$$
$$\implies y=b_ke^{r_kx}+b_{k-1}e^{r_{k-1}x}+\dots+b_1e^{r_1x}$$
where $r$ is a root of the equation
$$a_kr^k+a_{k-1}r^{k-1}+\dots+a_0=0$$
again assuming roots do not repeat.
A: Prime solutions for the Prouhet-Tarry-Escott problem of size ten ($10$ primes on the left side, other $10$ primes on the right side) .
$$ 2589701^k + 2972741^k + 6579701^k + 9388661^k + 9420581^k + 15740741^k + 15772661^k + 18581621^k + 22188581^k + 22571621^k $$
$$ = 2749301^k + 2781221^k + 6835061^k + 8399141^k + 10314341^k + 14846981^k + 16762181^k + 18326261^k + 22380101^k + 22412021^k $$
( Prime solution,    $ k = 1, 2, 3, 4, 5, 6, 7, 8, 9 $)
A: $$\prod_{n=1}^{\infty}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$$
Where $G$ is Catalan's constant.
A: An impressive relationship connecting three important functions in number theory
$$\sigma(n)+ \phi(n) = n \cdot d(n) \qquad \text{for } n \in \mathbb P $$
$\sigma(n)$ - sum of positive divisors of n
$\phi(n)$ - Euler's totient function
$d(n)$ - number of positive divisors of n
A: Simple and surprising
$$7! = 7\cdot{8}\cdot{9}\cdot{10}$$
$$\frac{1}{2}! = \sqrt{\frac{\pi}{4}}$$
$$\frac{123456789}{987654321} = 0.1249999988$$
A: •I will love to add this nested radical from Ramanujan:
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}=3$$ •Also like to add this too:$$1+2+3+4+\cdots=-\frac{1}{12}$$ •This famous Stirling'sapproximation$$n!\approx\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$$
A: $\mathrm{GCD}(F_{n},F_{m}) = F_{\mathrm{GCD}(n,m)}$ where $F_n$ is the $n$th Fibonacci number.
A: Another surprising equation
$$\bbox[7pt,border:3px #FF69B4 solid]{\color{red}{\large  213 \times 122 = 25986}}$$
Now read above expression in reverse order
$$\bbox[7pt,border:3px #FF69B4 solid]{\color{red}{\large  68952 = 221 \times 312}}$$
A: Personally I find this very interesting:
$$
\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}.
$$
A: I've found this to be rather surprising:


*

*$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=1$

*$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}=1$
As it essentially yields the identity:
$$\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}$$
It is surprising because obviously:
$$\forall{n\in\mathbb{N}}:\frac{1}{2^n}\neq\frac{1}{2^n\ln(2^n)}$$
In fact, the above inequity holds for every value of $n$, except for $n=\log_2e$.
Still, when summing up each of these infinite sequences, the result is $1$ in both cases.
A: Pfister's 16-Square Identity:
$$(x_1^2+x_2^2+x_3^2+\dots+x_{16}^2)(y_1^2+y_2^2+y_3^2+\dots+y_{16}^2) = z_1^2+z_2^2+z_3^2+\dots+z_{16}^2$$
where the $z_i$ are rational functions of the $x_i, y_i$. One would have thought that $n$ square identities are only for $n = 1,2,4,8$, but non-bilinear ones in fact are for all $n = 2^m$.
A: A rather simple one $$2^4 = 4^2$$
You can use this one to "proof" (as a prank) that $x^y = y^x$
A: It is still strange for me
$$
i^i = e^{\pi(2k-\frac{1}{2})}.
$$
And so, one could say $i^i\in\mathbb R$.

Note that $i^i$ is a sequence of real numbers and actually $i^i\not\in\mathbb R$, but still $i^i\subset\mathbb R$.
A: Taken from the first question I posed upon joining M.SE:
Define a function $f(\alpha, \beta)$, $\alpha \in (-1,1)$, $\beta \in (-1,1)$ as
$$ f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$
You can use, for example, the Residue Theorem to show that
$$ f(\alpha, \beta) = \frac{\pi \sin{\pi \alpha \beta}}{ \sin{\pi \alpha}  \sin{\pi \beta}} $$
Clearly, from this latter expression, $f(\alpha, \beta) = f(\beta, \alpha)$.  But from where does such a symmetric result come?  The integral itself does not lend itself to predicting any such symmetry so far as I (and many others so far) can see.
A: $$\sum_{n=1}^\infty(n\,\operatorname{arccot}n-1)=\frac12+\frac{17\pi}{24}-\ln\sqrt{e^{2\pi}-1}+\frac1{4\pi}\operatorname{Li}_2\left(e^{-2\pi}\right),$$
where $\operatorname{Li}_2$ is the dilogarithm.
A: Using the mystical ennead to calculate the decimal expressions for fractions of 7. Start with this figure:

Then follow the connected path, giving the sequence 1 4 2 8 5 7. Then you write this sequence starting on each digit, in order, giving
. 1 4 2 8 5 7 = 1/7
. 2 8 5 7 1 4 = 2/7
. 4 2 8 5 7 1 = 3/7
. 5 7 1 4 2 8 = 4/7
. 7 1 4 2 8 5 = 5/7
. 8 5 7 1 4 2 = 6/7
A: $$(1+2+3+\cdots+n)!=1!3!5!\cdots(2n-1)!$$for $n=0,1,2,3,4$.
A: If $x^n + y^n + z^n=0$ and $xyz \ne 0$, then
$$\frac{(x^n-y^n)^2}{(xy)^n} + \frac{(y^n-z^n)^2}{(yz)^n} + \frac{(z^n-x^n)^2}{(zx)^n} = -9.$$
A: $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
was surprising to me when I saw it for the first time.
A: $$10^2+11^2+12^2=13^2+14^2$$ I found that one stunning.
P.S. In general, for $n>0$, the sum of $n+1$ consecutive squares starting with $x_1 = 2n^2+n$ is equal to $n$ consecutive squares starting with $y_1 = x_1+(n+1)$. Hence,
$$3^2+4^2 = 5^2$$
$$10^2+11^2+12^2=13^2+14^2$$
$$21^2+22^2+23^2+24^2 = 25^2+26^2+27^2$$
and so on.
A: By far my favorite identity: $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$
The fun part about this one (for me) is that it looks absolutely false at first glance.  They both evaluate to $\pi$.
A: One of the most worth-mentioning identities may probably be Carl Friedrich Gauss's compution of $\cos(\frac{2\pi}{17})$:
\begin{align}
\cos(\frac{2\pi}{17})&=
\frac1{16}[-1+\sqrt{17} + \sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}]\\
\end{align}
which has a significant role in regular heptadecagon construction.
A: This is slightly contrived, but consider a situation where you have two balls, of mass $M$ and $m$, where $M=16\times100^N\times m$ for some integer $N$.  The balls are placed against a wall as shown: 

We push the heavy ball towards the lighter one and the wall.  The balls are assumed to collide elastically with the wall and with each other.  The smaller ball bounces off the larger ball, hits the wall and bounces back.  At this point there are two possible solutions: the balls collide with each other infinitely many times until the larger ball reaches the wall (assume they have no size), or the collisions from the smaller ball eventually cause the larger ball to turn around and start heading in the other direction - away from the wall.  
In fact, it is the second scenario which occurs: the larger ball eventually heads away from the wall.  Denote by $p(N)$ the number of collisions between the two balls before the larger one changes direction, and gaze in astonishment at the values of $p(N)$ for various $N$: 
\begin{align}
p(0)&=3\\
p(1)&=31\\
p(2)&=314\\
p(3)&=3141\\
p(4)&=31415\\
p(5)&=314159\\
\end{align}
and so on.  $p(N)$ is the first $N+1$ digits of $\pi$!  
This can be made to work in other bases in the obvious way.  
See 'Playing Pool with $\pi$' by Gregory Galperin. 
A: $$\cos \left(20\right) \cos \left(40\right) \cos \left(80\right) = \frac{1}{8}$$
for angles in degrees.  This identity is interesting for its historical association with the teenage Richard Feynman.  From Genius by James Gleick:
"He and his friends traded mathematical tidbits like baseball cards.  If a boy named Morrie Jacobs told him that the cosine of 20 degrees multiplied by the cosine of 40 degrees multiplied by the cosine of 80 degrees equaled exactly one-eighth, he would remember that curiosity for the rest of his life, and he would remember that he was standing in Morrie's father's leather shop when he heard it."
A: There are many fantastic equations that I've seen, but the one that definitely sticks out as top in my mind is the Atiyah-Singer Index theorem (it can be written in many ways; this is the way that I first learned it).
$$\operatorname{Ind}(D) = (-1)^n \int_M \frac{\operatorname{ch}(E)-\operatorname{ch}(F)}{\operatorname{e}(TM)} \operatorname{Td}^{hol}(TM \otimes \mathbb C)$$
Here $M$ is a $2n$-dimensional smooth compact manifold, with $E$, $F$ vector bundles over $M$, $D:\Omega^0(E) \rightarrow \Omega^0(F)$ is an elliptic differential operator, $\operatorname{Ind}(D) = \dim \ker D - \dim \operatorname{coker} D$, $\operatorname{ch}$ denotes the Chern class, $\operatorname{e}$ is the Euler class, and $\operatorname{Td}^{hol}$ is the holomorphic Todd class. Note that in this formulation the choice of the divisor depends naturally on the choice of $D$, a fact which is obscured in the notation.
It's probably the only equation I've ever seen in math which forced me to think about two entire fields in a different way. The mere fact that something like this connecting analysis and topology at such a deep level could be true is really incredible.
A: This one is no less-
Let $d$ be the distance between Incenter($r$) and Circumcenter ($R$) Then-
      $$R^2-d^2=2Rr$$ and this one $$\frac{1}{R-d}+ \frac{1}{R+d}=\frac{1}{r}$$
A: $$(1+i)(1+2i)(1+3i) = (1-i)(1-2i)(1-3i)$$
A: The series $$\sum_{n=1}^{\infty} \frac{n^{13}}{e^{2\pi n} - 1} = \frac{1}{24}$$ is not entirely obvious. (At this time WolframAlpha is unable to find its closed form.)
A: $3^3 + 4^3 + 5^3 = 6^3$.
Also,
$1/89 = 0.01 + 0.001 + 0.0002 + 0.00003 + 0.000005 + 0.0000008 + 0.00000013 + \cdots$.

Let $S = \sum \frac{F_n} {k^n}$. Then $S + kS =  1 + \sum \frac{ F_{n} + F_{n-1} } {k^n} = 1 + \sum \frac {F_{n+1}}{k^n} = 1 + k^2S -1 - k$
In particular, for $k=10$, we get $ S = \frac{10}{89}$. Divide by 10 to get the second equation.
A: $$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac4{15}\right)}{\Gamma\left(\frac13\right)\Gamma\left(\frac2{15}\right)}=\frac{\sqrt2\,\,\sqrt[20]3}{\sqrt[6]5\,\sqrt[4]{5-\frac{7}{\sqrt{5}}+\sqrt{6-\frac{6}{\sqrt{5}}}}}=\frac{\phi \,\,  \sqrt[20]3 \,\, \sqrt{\!\sqrt 3 \cdot \sqrt[4] 5-\phi^{3/2}}}{\sqrt 2 \,\, \sqrt[24] 5}$$
A: from $1$ years ago 
$$\tan x=\cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-...}}}}$$
A: Something very exotic... and I do not know, whether this example fits the bill for this question here. But let's see. 
In some sense it seems to be possible to assign equality
$$  e = \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}\tfrac 1{e^8}...$$

Originally I thought I had a mathematical contradiction when I wrote:
$$ \text{ How can } \qquad e^{-1-2-4-8-16-...} = e^{-1/(1-2)}= e^{+1} = e \qquad \text{?}$$

Initially I thought that this were an example where the rule of the closed form of the geometric series might break. The equality seems impossible because the product is even of only decreasing factors and should so be convergent and moreover converge to zero:
$$e^{-(1+2+4+8+16+...)} = \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}... $$
 so 
$$  e \overset{???}{=} \tfrac 1{e^1} \tfrac 1{e^2} \tfrac 1{e^4}\tfrac 1{e^8}...$$
Well, in some circumstances in the context of divergent series we observe strange things - but here the factors are all nicely decreasing and no unexpected effect should occur.
But - that this actually holds in some sense was mentioned by Robert Israel here in MSE
A: I believe that one that should be mentioned is the prime number theorem:
 Let $\pi(x)$ be the number of primes not exceeding $x$. Then:
$\pi(x)\sim \frac{x}{logx}$
A: Wheatstone's identity, which shows how powers can be constructed via arithmetic progressions:
$$n^a = \sum_{k=0}^{t-1}\biggl(\frac{n^a}{t}-\frac{\delta(t-1)}{2}+k\delta\biggr).$$
Put into words: An arithmetic progression of $t$ terms with constant difference $\delta$ and first term $\tfrac{1}{t}n^a-\tfrac{1}{2}\delta(t-1)$ will sum to $n^a$.
A: Not sure if it's an equation but.. I was very surprised that among all two-dimensional compact orientable surfaces, a 2-sphere $S^2$ is the only one that has nontrivial higher homotopy groups. So, from the point of view of homotopy, sphere is the "most complicated" surface in some sense.
A: I have found the identity below from which can be deduced infinitely many others of increasing degree by the elementary theory of elliptic curves. 
The 6-tuples indicate the coefficients of, respectively $n^5$, $n^4$, $n^3$, $n^2$, $n$ and $1$
$$ A = (1, 10, -8, 16, 64, -32) \\
 B = (1, -10, -8, -16, 64, 32) \\
 C = (-1, 8, 8, -16, 80, 32) \\
 D = (-1, -8, 8, 16, 80, -32)$$
Then it is verified the identity
 $$n(6n^4 +24n^2 + 96)^3 = A^3 + B^3 + C^3 + D^3$$
The factor of $n$ at the left is never nul for any integer $n$.
A: For the orthogonal ($\vec{v}_\perp$) and parallel component ($\vec{v}_\parallel$) of a $\Bbb R^3$ vector $\vec{v}$ along another vector $\vec{k}$ with lenght $k$ we have
$$ k\,\vec{v}\, k= \underbrace{\vec{k}\,(\vec{v}\cdot\vec{k})}_{k \vec{v}_\parallel k} + \underbrace{\vec{k}\times(\vec{v}\times\vec{k})}_{k \vec{v}_\perp k} $$
or a bit less showy
$$\vec{v}= \frac{\vec{k}\,(\vec{v}\cdot\vec{k}) + \vec{k}\times(\vec{v}\times\vec{k})}{k^2},$$
which for unit length vectors $\vec{k}$ gives obvioulsy
$$ \vec{v}= \underbrace{\vec{k}\,(\vec{v}\cdot\vec{k})}_{\vec{v}_\parallel} + \underbrace{\vec{k}\times(\vec{v}\times\vec{k})}_{\vec{v}_\perp}. $$
To me this looked a first quite surprising since the scalar, the cross and the product of scalar and vector, are combined in a rather symmetric and harmonic fashion here. This is kind of unexpected, since as we know the one is a vector and the other one is a scalar. Though its of course very easy to show that this is correct and also the one contains the $\sin$ and the other the $\cos$ of the angle in between $\vec{k}$ and $\vec{v}$. But its more the vector algebra which kind of amazed me.
A: The formulas obtained by Robert Scheider and involving The Golden Ratio $\phi$, the Euler Totient $\varphi (n)$
 and the Moebius Function $\mu(n)$, are rather surprising
$$
\begin{array}{*{20}c}
   {\phi  =  - \sum\limits_{1\, \le \,k} {\frac{{\varphi (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} } & {e^{\,\phi }
  = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\varphi (k)}}{k}} } }  \\
   {\frac{1}{\phi } =  - \sum\limits_{1\, \le \,k} {\frac{{\mu (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} } & \begin{array}{l}
 e^{\,{{1\,} \mathord{\left/
 {\vphantom {{1\,} {\,\phi }}} \right.
 } {\,\phi }}}  = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\mu (k)}}{k}} }  =  \\ 
  = \frac{1}{e}\prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\, - \;\frac{{\varphi (k)}}{k}} }  \\ 
 \end{array}  \\
   {1 = \sum\limits_{1\, \le \,k} {\frac{{\mu (k) - \varphi (k)}}{k}\ln \left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)} }
 & {e = \prod\limits_{1\, \le \,k} {\left( {1 - \frac{1}{{\phi ^{\,k} }}} \right)^{\,\frac{{\mu (k) - \varphi (k)}}{k}} } }  \\
\end{array}
$$
Also refer to this Wikipedia article.
A: A fun one from probability: If $U, V, W$ are independent random variables, each uniform on $(0,1)$, then $(UV)^W$ is also uniform on $(0,1)$.
A: Further working on 
TobiMcNamobi 's identity:
$$
\begin{array}{rcl}
\dfrac{1}{7} & = & \sum\limits_{k=1}^{\infty} 2^k\times7\times10^{-2k} \\
\dfrac{1}{49} & = & \sum\limits_{k=1}^{\infty} 2^k\times10^{-2k} \\
& = & \sum\limits_{k=1}^{\infty} 2^k\times100^{-k} \\
& = & \sum\limits_{k=1}^{\infty} \left(\dfrac{2}{100}\right)^k \\
& = & \sum\limits_{k=1}^{\infty} \left(\dfrac{1}{50}\right)^k
\end{array} \\ \\
\boxed{\dfrac{1}{49} = \dfrac{1}{50} + \dfrac{1}{2500} + \dfrac{1}{625000} + \cdots}
$$
A: I was amazed at one of the results that Ramanujan sent to Hardy
in his first letter.

$$ (1.10) \text{ If }
 u = \frac{x}{1+} \frac{x^5}{1+} \frac{x^{10}}{1+}
\frac{x^{15}}{1+\dots},\, v = \frac{x^{1/5}}{1+}
 \frac{x}{1+}\frac{x^{2}}{1+}\frac{x^{3}}{1+\dots},\\
\text{then}\qquad\qquad v^5 = u\frac{1-2u+4u^2-3u^3+u^4}
{1+3u+4u^2+2u^3+u^4}
$$

This is just one of the modular equations satisfied by the
Rogers-Ramanujan continued fraction.
The Wikipedia article mentions other amazing facts about the
continued fraction but this result stands out for me.
A: Perhaps this is mainly for the programmers among us, a few fun 'cyclical' identities between bases $10$ and $16$:
$$8200_{10}=2008_{16}$$
$$4100_{10}=1004_{16}$$
$$1041_{10}=0411_{16}$$
In case you're confused, the rule for the above is:
$$\mathrm{CycleDigitsLeft}(N_{10},1)=N_{16}$$
I thought it was cute :)
