How does one get that $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$? While watching interesting mathematics videos, I found one of the papers  of Srinivasa Ramanujan to G.H.Hardy in which he had written $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$.
The problem is that every term on the left is more than $\frac{1}{120}$ yet the sum is $\frac{1}{120}$. How is that ???
I know that there are many and much more interesting things  presented by Ramanujan (like $1-1+1-1+1...=\frac{1}{2}$ and $1+2+3+4.....=\frac{-1}{12}$) but for now I am interested in the summation in the title.
Any idea/hint is heartily welcome. Thanks.
Here is the video I'm talking about.
 A: Let $f(x)=x^d$ for $d\geq 1$ then by Euler–Maclaurin summation formula 
$$\sum_{k=1}^nf(x)=C+\int_0^n f(x) dx+\frac{f(n)}{2}+\sum_{k\geq 1}\frac{B_{2k}}{(2k)!}f^{(2k-1)}(n)$$
where $B_n$ denotes the $n$-th Bernoulli number.
In his theory of Divergent Series,  Ramanujan "identified" $\sum_{k=1}^{\infty}f(x)$ with $C$. Note that $C$ is zero if $d$ is even and it is $-\frac{B_{d+1}}{d+1}$ if $d$ is odd. So for $d=1$ we get $-\frac{1}{12}$ and for $d=3$ we obtain $\frac{1}{120}$.
A: Below is the derivation of the following equality:

$$\zeta(-s)=\frac1{1-2^{1+s}}\lim_{r\to1^-}\left(\underbrace{r\cdot\frac d{dr}r\cdot\frac d{dr}\dots r\cdot\frac d{dr}}_s\frac{-1}{1+r}\right)$$


Notice that if we have
$$\zeta(s)=1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\dots$$
$$\eta(s)=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\dots$$
then, when they are absolutely convergent, we get the functional equation:
$$\begin{align}\zeta(s)&=1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\dots\\\eta(s)&=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\dots\\\hline\zeta(s)-\eta(s)&=2\left(\ \ \frac1{2^s}\qquad\ \ +\frac1{4^s}+\dots\right)\\&=2^{1-s}\left(1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\dots\right)\tag{factored out $2^s$}\\&=2^{1-s}\zeta(s)\end{align}$$
Solving for $\zeta(s)$, we get
$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)$$
And since these are analytic functions, this holds for all $s$.
As a somewhat elementary method, we can evaluate the $\eta(s)$ as follows, which converges for all $s$.
$$\eta(s)=\lim_{r\to1^-}\sum_{n=1}^\infty\frac{(-1)^{n+1}r^n}{n^s}$$
And thus, we have ourselves left with
$$\eta(-3)=\lim_{r\to1^-}\sum_{n=1}^\infty(-1)^{n+1}r^nn^3=-\frac18$$
which may be done by manipulating the geometric series:
$$\frac{-1}{1+r}=\sum_{n=0}^\infty(-1)^{n+1}r^n$$
$$r\cdot\frac d{dr}\frac{-1}{1+r}=\sum_{n=0}^\infty(-1)^{n+1}nr^n\\\vdots$$
$$\underbrace{r\cdot\frac d{dr}r\cdot\frac d{dr}\dots r\cdot\frac d{dr}}_k\frac{-1}{1+r}=\sum_{n=0}^\infty(-1)^{n+1}n^kr^n$$
Which then yields the desired result $\zeta(-3)=\frac1{120}$.
So, succinctly, we end up with the following form for the zeta function whenever $s\in\mathbb N,\ s>0$:

$$\zeta(-s)=\frac1{1-2^{1+s}}\lim_{r\to1^-}\left(\underbrace{r\cdot\frac d{dr}r\cdot\frac d{dr}\dots r\cdot\frac d{dr}}_s\frac{-1}{1+r}\right)$$

which isn't terribly difficult to work out.  Just a lot of quotient rule going on in there.  If $s=0$, we have
$$\zeta(0)=\frac1{1-2^{1+0}}\lim_{r\to1^-}\left(\frac1{1+r}\right)=-\frac12$$

Some values worked out for you:
$$\zeta(-1)=\frac1{1-2^{1+1}}\lim_{r\to1^-}\left(r\cdot\frac d{dr}\frac{-1}{1+r}\right)=-\frac13\lim_{r\to1^-}\left(\frac{r}{(1+r)^2}\right)=-\frac1{12}$$
$$\zeta(-2)=\frac1{1-2^{1+2}}\lim_{r\to1^-}\left(r\cdot\frac d{dr}r\cdot\frac d{dr}\frac{-1}{1+r}\right)=-\frac17\lim_{r\to1^-}\left(r\cdot\frac d{dr}\frac{r}{(1+r)^2}\right)\\=-\frac17\lim_{r\to1^-}\left(\frac{r-r^2}{(1+r)^3}\right)=0$$
$$\zeta(-3)=\frac1{1-2^{1+3}}\lim_{r\to1^-}\left(r\cdot\frac d{dr}r\cdot\frac d{dr}r\cdot\frac d{dr}\frac{-1}{1+r}\right)=-\frac1{15}\lim_{r\to1^-}\left(r\cdot\frac d{dr}\frac{r-r^2}{(1+r)^3}\right)\\=-\frac1{15}\left(\frac{r-4r^2+r^3}{(1+r)^4}\right)=\frac1{120}$$
A: 
The problem is that every term on the left is more than $\frac{1}{120}$ yet the sum is $\frac{1}{120}$. How is that ???

To answer the question directly: Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series. It is well explained in the linked Wikipedia article. 
A: I think I have rediscovered it (after watching the video you linked and read wiki biography of Ramanujan). 
Start with $$
\frac{1}{x+1} =1 -x +x^2 -x^3 +-\cdots, \quad |x| <1.
$$ and differentiate to get $$
-\frac{1}{(x+1)^2} =-1 +2x -3x^2 +4x^3 -+\cdots, \quad |x| <1. \\
\frac{2}{(x+1)^3} =2 \cdot 1 -3 \cdot 2x +4 \cdot 3 x^2 -+\cdots, \quad |x| <1. \\
-\frac{6}{(x+1)^4} =-3 \cdot 2 \cdot 1 +4 \cdot 3 \cdot 2x -5 \cdot 4 \cdot 3 x^2 +-\cdots, \quad |x| <1.
$$ 
Take a magic mushroom and, ignoring $|x| <1$, let us take $x=1$ in each one. $$
\frac{1}{2} =1 -1 +1 -1 +- \cdots \\
-\frac{1}{4} =-1 +2 -3 +4 -+ \cdots \\
\frac{1}{4} =2 \cdot 1 -3 \cdot 2 +4 \cdot 3 -+ \cdots \\
-\frac{3}{8} =-3 \cdot 2 \cdot 1 +4 \cdot 3 \cdot 2 -5 \cdot 4 \cdot 3 +- \cdots
$$ Or more formally, $$
\sum_{m=1}^\infty (-1)^{m+1} m =\frac{1}{4}. \\
\sum_{m=1}^\infty (-1)^{m+1} m (m+1) =\frac{1}{4}. \\
\sum_{m=1}^\infty (-1)^{m+1} m (m+1) (m+2) =\frac{3}{8}.
$$
But notice $$ \begin{align}
\sum_{m=1}^\infty (-1)^{m+1} m^2
=&\sum_{m=1}^\infty (-1)^{m+1} m (m+1) -\sum_{m=1}^\infty (-1)^{m+1} m \\
=&\frac{1}{4} -\frac{1}{4} =0.
\end{align}
$$ and $$ \begin{align}
\sum_{m=1}^\infty (-1)^{m+1} m^3
=&\sum_{m=1}^\infty (-1)^{m+1} m (m+1) (m+2) -3\sum_{m=1}^\infty (-1)^{m+1} m^2 -2\sum_{m=1}^\infty (-1)^{m+1} m \\
=&\frac{3}{8} -3 \cdot 0 -2 \cdot \frac{1}{4}
=-\frac{1}{8} \quad \quad \ldots \spadesuit
\end{align}
$$
On the other hand, $$
\zeta(-3) :=1^3 +2^3 +3^3 +\cdots \\
2^4 \zeta(-3) =2 \cdot 2^3 +2 \cdot 4^3 +2 \cdot 6^3 +\cdots \\
$$ Subtract them, aligning the 2nd, 4th, 6th term like Ramanjunan did in his notebooks (shown in the video). $$
-15 \zeta(-3) =1^3 -2^3 +3^3 -+\cdots \quad \quad \ldots \heartsuit
$$ $\heartsuit$ and $\spadesuit$ together give us: (Hold your breath.) $$
\sum_{m=1}^\infty m^3 =\frac{1}{120}.
$$
Recently I also found a proof of Riemann conjecture, but the answer box is too narrow for me to type all that down. 
P.s. seriously, I think Ramanujan's effort is sort of finding an interpretation of divergent series so that they have a real value, while their manipulation to be still consistent to our usual notion of arithmetics: arranging, addition, expanding, etc.? 
Maybe this can be compared to the attempt to define quaternion as an extension of complex numbers, while inevitably discarding commutative law?
A: When you define a series like  $1^3+2^3+3^3+4^3+...+n^3$ you have a specific definition in mind for How to compute this series and How to take the limit as $n$ tends to infinity.
Ramanujan is using a different definition for what it means to compute this series. That definition allows for properties like the sum of numbers being equal to a value smaller than every term. 
