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136

A variant: note that $$\color{red}{\mathbf 1}=0+\color{red}{\mathbf 1}=0+0+\color{red}{\mathbf 1}=0+0+\cdots+0+\color{red}{\mathbf 1}=0+0+0+\cdots$$ and $$\color{green}{\mathbf 2}=0+\color{green}{\mathbf 2}=0+0+\color{green}{\mathbf 2}=0+0+\cdots+0+\color{\mathbf green}{2}=0+0+0+\cdots$$ "Since the right hand sides are the same", this proves that ...


48

Another example where dots are misleading: $$1= \frac{ 1 \cdot \color{blue}{2} \cdot \color{green}{3} \cdot \color{red}{4} \cdots}{ 2 \cdot \color{blue}{3} \cdot \color{green}{4} \cdot \color{red}{5} \cdots} \leq \frac{1}{2}$$


21

As I've noted in the comments, the Śleszyński-Pringsheim Theorem guarantees that this continued fraction converges to a finite value (since the prime numbers are always greater than or equal to $2$). This arXiv preprint gives an approximate value $s$ for the continued fraction whose partial denominators are prime numbers: ...


21

Euler proved in "De Transformatione Serium in Fractiones Continuas" Reference: The Euler Archive, Index number E593 (On the Transformation of Infinite Series to Continued Fractions) [Theorem VI, §40 to §42] that $$s=\cfrac{1}{1+\cfrac{2}{2+\cfrac{3}{3+\cdots }}}=\dfrac{1}{e-1}.$$ Here his an explanation on how he proceeded. He stated that if ...


20

Here's a nice little Mathematica routine for evaluating Tito's continued fraction with precision prec: prec = 10^4; y = N[4, prec]; c = y; d = 0; k = 1; u = 1; v = y; While[True, c = 1 + u/c; d = 1/(1 + u d); h = c*d; y *= h; v += 96 k^2 + 8; c = v + u/c; d = 1/(v + u d); h = c*d; y *= h; If[Abs[h - 1] <= 10^-prec, Break[]]; u += 3 k (k + 1) + 1; ...


20

The first expression is a continued fraction, the second isn't. A continued fraction is the limit of $$ a_0, a_0 + \frac{1}{a_1}, a_0 + \frac{1}{a_1 + \frac{1}{a_2}} \ldots $$ for a fixed sequence of natural numbers $a_0, a_1, a_2 \ldots$ The second expression is the limit of fractions which look similar to these fractions, but which don't correspond to ...


19

This is of the same type as the following $$0=(1-1)+(1-1)+(1-1)+\ldots=1+(-1+1)+(-1+1)+(-1+1)+\ldots = 1 \; .$$ Did's example is even simpler and closer to yours in spirit. In working with an infinite number of operations, you have to be very careful about how you are performing them. Normally, one uses some kind of limit, but then what you really do is ...


18

Why don't you consider that $$ \arctan x + \arctan \frac{1}{x}=\frac{\pi}{2}, $$ so that, for large values of $x>0$, $$ \arctan x = \frac{\pi}{2}-\arctan {1 \over x} = \frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^3}-\frac{1}{5x^5} +\ldots ? $$ See also this discussion.


17

From the definition, you get $$\frac{x}{1+f(x)}=f(x)$$ Thus $$x=(1+f(x))f(x)$$ Now differentiate $$1=f'(x)+2f(x)f'(x)=f'(x)(1+2f(x))$$ Or $$f'(x)=\frac{1}{1+2f(x)}$$ Now, $f(0)=0$, from definition of $f$. One remark though: I did not prove the continued fraction is differentiable or even convergent. I just assume it's true, then you can compute ...


17

As with Mariano, I too have no idea what sort of mathematical sorcery Ramanujan used, but what you have there is a special case of the Rogers-Ramanujan continued fraction (sans the $\sqrt[5]{q}$ factor). Letting $$R(q)=\cfrac{\sqrt[5]{q}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cdots}}}$$ the question amounts to how one might arrive at ...


16

It might be not a direct answer to your question, but it is possible that there is no 2nd term of the continued fraction in question. I believe it is a long-standing open problem if $\,{^5 e}\in\mathbb{N}$, and, in general, for every integer $n \ge 5$, if $\,{^n e}\in\mathbb{N}$ (and also, for every integer $n \ge 4$, if $\,{^n \pi}\in\mathbb{N}$). It is ...


16

This is a general feature of continued fractions. What you are doing is iterating the function $$ s(x) = \frac{1}{2-x} $$ ... so, for instance, you have that $$ 1 = s(1) = s(s(1)) = s(s(s(1))) = \cdots $$ and $$ 2 = s(3/2) = s(s(4/3)) = s(s(s(5/4))) = \cdots .$$ The function $s$ has a single fixed point at 1 (first line). It is known that iterating $s$ ...


15

Hint: If $a = \frac{1}{1+\frac{1}{1+\cdots}}$ then $\frac{1}{a}-1 = \frac{1}{1+\frac{1}{1+\cdots}} = a$. Can you take it from here?


15

The iterated integral of the complementary error function, $$\begin{align*} \mathrm{i}^n\mathrm{erfc}(z)&=\underbrace{\int_z^\infty\int_{t_{n-1}}^\infty\cdots\int_{t_1}^\infty}_{n} \mathrm{erfc}(t)\,\mathrm dt\cdots\mathrm dt_{n-2}\mathrm dt_{n-1}\\ &=\frac2{n!\sqrt\pi}\int_z^\infty(t-z)^n\exp(-t^2)\,\mathrm dt \end{align*}$$ (see e.g. Abramowitz ...


15

This isn't exceptionally good compared to the partial convergents of the continued fraction expansion. Terminating the continued fraction for $\pi$ right before the $292$ gives $\frac{355}{113}= \textbf{3.141592}9203\ldots$, which gets $6$ digits after the decimal place right, while your fraction only gets $3$ digits after the decimal place correct even ...


14

There is a lot of motivation behind this identity (although I don't know how it relates to Ramanujan's motivation, if at all). The function $R(q)$ described in J.M.'s answer is an example of a modular function (see the wikipedia entry), and the general theory of complex multiplication says that if $\tau$ is a quadratic irrational algebraic number (e.g. $i$) ...


14

The first one is expressible in terms of the modified Bessel function of the first kind: $$1+\cfrac1{2+\cfrac1{3+\cfrac1{4+\cdots}}}=\frac{I_0(2)}{I_1(2)}=1.433127426722\dots$$ The second one, through an equivalence transformation, can be converted into the following form: $$1+\cfrac1{\frac12+\cfrac1{\frac23+\cfrac1{\frac38+\cfrac1{b_4+\cdots}}}}$$ where ...


13

This transformation of a series into its equivalent continued fraction, with the series partial sums being equal to the continued fraction convergents, is due to Euler. The series $$\sum_{n\geq 0}c_{n}=c_0+c_1+\dots+c_n+\dots$$ is transformed into the continued fraction $$b_0+\mathbf{K}\left( a_{n}|b_{n}\right) =b_0+\dfrac{a_{1|}}{|b_{1}}+\dfrac{a_{2}|}{% ...


11

$$a_{n+1} = \frac{1}{1+a_n},\ a_0=1/x,\ \sum_{n=0}^{\infty}a_n=1$$ Unfortunately, $a_n$ converges to a non zero constant $\frac{\sqrt{5}-1}{2}$, so that the sum of all $a_n$'s is infinity.


10

If the continued fraction converged to $\pi$, the value of the odd convergent $$ 3 +\cfrac1{ 5 +\cfrac1{ 7}} = \frac{115}{36} = 3.19444\ldots $$ would be strictly smaller than $\pi$. But the convergent is already too big. The actual value of the continued fraction must exceed $\frac{115}{36}$; it lies somewhere between $\frac{115}{36} \approx 3.1944$ ...


10

After looking at my previous hint, I was unable to proceed as easily as I thought. Instead, I have here expanded the division of power series in detail. Start with $$ \begin{align} \frac{\cos(x)}{\sin(x)/x} &=\frac{\sum\limits_{k=0}^\infty(-x^2)^k\dfrac1{(2k)!}} {\sum\limits_{k=0}^\infty(-x^2)^k\dfrac1{(2k+1)!}}\\[12pt] ...


10

We deal with a function $x\mapsto \frac{1}{1+\frac{1}{x}}$, which if we take $\frac{1}{x}$ as argument, we should rather write as $x\mapsto \frac{1}{1+x}$. Iterated, this gives the fixed point iteration for finding a soltion of $\frac{1}{1+x}=x$. And therefore all your terms eventually turn out converge to a solution of $1=(1+x)\ x$, namely ...


10

An expansion of Andre's comment into a detailed exposition can be found at http://www.johnderbyshire.com/Opinions/Diaries/Puzzles/2009-06.html. It's also at http://www.angelfire.com/ak/ashoksandhya/winners2.html#PUZZANS4.


10

We can deduce both of (1) and (2), and the fact that $p$ is a sum of two squares, by playing with quadratic forms. I have the impression this is all "well known to those who know it well". But I've never seen this elementary argument written down in elementary language, so here it is. We begin with some general discussion of quadratic equations. Let the ...


9

Found why it's $-1$. Rewrite the equation as :$$n\bigg(\frac{1+\frac1n\sqrt{n^2 + 4n}}{2}\bigg)$$ The square root can be written as $\sqrt{n^2(1+4/n)} = |n|\sqrt{1+4/n} = -n\sqrt{1+4/n}$ because $n<0$. Then you obtain : \begin{eqnarray*} \phi(n) &=& \frac{n}{2}\big(1-\sqrt{1+4/n}\big)\\ &=& \frac n2(1-(1+\frac2n) + O(n^{-2}))\\ ...


9

Here is how you find the continued fraction for any number at all. Say the number is $x_0$. First, let $a_0$ be the largest integer that does not exceed $x_0$. That is, $a_0 = \lfloor x_0\rfloor$. And let $b_0$ be the fractional part of $x_0$, that is $b_0 = x_0 - a_0$. In our example, $$\begin{align}x_0&=\pi,\\ a_0 &= 3,\\ b_0 &= ...


9

Suppose that the original (irrational) number has the continued fraction $[1;x,y,\ldots]$. After one iteration, the number is of the form $$1+\frac{1}{1+1/x+\frac{1}{1+1/(x+1/y)+\epsilon}},$$ where $\epsilon < 1$. Its continued fraction starts $[1;z,w,\ldots]$, where $z$ is the floor of $$1+1/x+\frac{1}{1+1/(x+1/y)+\epsilon}.$$ It's easy to see that this ...


9

We use the formula given here: Gauss' continued fraction for $\tan z$ and see that $$\tan(1) = \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5 -\dots}}}$$ Now use the identity $$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$ To transform $$\cfrac{1}{a - \cfrac{1}{b - \cfrac{1}{c - \dots}}}$$ to $$\cfrac{1}{a-1 + \cfrac{1}{1 + ...


9

The question seems to be a bit confused about whether $n\lfloor a\rfloor + 1 \leq \lfloor an \rfloor$ is supposed to hold for all $n$, for all $n \geq N$ where $N$ is chosen independent from $a$, or for all $n \geq N_a$ (i.e., what is sufficiently large may depend on $a$). In the latter case, the following shows that the inequality is true for all $a \in ...



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