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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 ...

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: ...

20

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; ...

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 ...

17

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

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

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 ...

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 ...

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 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

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 ...

15

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 ...

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 $$b_0+\mathbf{K}\left( a_{n}|b_{n}\right) =b_0+\dfrac{a_{1|}}{|b_{1}}+\dfrac{a_{2}|}{% ... 10 We deal with a function x\mapsto \frac{1}{1+\frac{1}{x}}, which if we take \frac{1}{x} as argument, we should rather write as x\mapsto \frac{1}{1+x}. Iterated, this gives the fixed point iteration for finding a soltion of \frac{1}{1+x}=x. And therefore all your terms eventually turn out converge to a solution of 1=(1+x)\ x, namely ... 10 After looking at my previous hint, I was unable to proceed as easily as I thought. Instead, I have here expanded the division of power series in detail. Start with$$ \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

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 $\frac{115}{36} \approx 3.1944$ ...

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 $a \in ... 9 Suppose that the original (irrational) number has the continued fraction$[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 $$\cfrac{1}{a-1 + \cfrac{1}{1 + ... 9 An expansion of Andre's comment into a detailed exposition can be found at http://www.johnderbyshire.com/Opinions/Diaries/Puzzles/2009-06.html. It's also at http://www.angelfire.com/ak/ashoksandhya/winners2.html#PUZZANS4. 8 You can find the proofs here, taken from Chrystal's Algebra - which is one of the best references on continued fractions. I suspect that if you read Chrystal then almost all of your questions will be answered. See also chapter 9 of Fowler's "The mathematics of Plato's Academy" 8 You have shown that the sequence of a_{2n} converges. Call e (for even) its limit, and call o the limit of a_{2n+1}. Then the recursive relation giving a_{n+1} in terms of a_n implies that$$e=1+1/(1+o)$$and$$o=1+1/(1+e),$$from which it follows that e=o=\sqrt2. [For example, take the limit as n\to\infty of a_{2n}=1+1/(1+a_{2n-1}) to ... 8 First hint: It’s 1/φ. Check out the golden ratio. Second hint, which adds a bit to answers given so far: Since you should probably apply limits formally, you can describe it as the limit of a sequence given by a_0 = 1 and a_{k+1} = \frac{1}{1 + a_k}. Now if x = \lim_{k → ∞} a_k, then$$\frac{1}{1+x} \overset{\text{limit rules}}{=} \lim_{k → ∞} ... 8 Consider the more general continued fraction $$\cfrac{1}{x - \cfrac{1}{3x- \cfrac{1}{5x- \cfrac{1}{7x - \cfrac{1}{9x - \ddots}}}}}$$ Some numerical experimentation suggests that this is equal to$\tan (1/x)$. Then we can substitute$x = \frac{5}{2 \pi}$to get the original identity. Edit: This identity is the third continued fraction identity for tangent ... 7 No, continued faction expansions repeat only for the so-called "quadratic irrationals", numbers$x$satisfying$ax^2 + bx + c = 0$for some integers$a, b$and$c$. In particular, the continued fraction expansion for the algebraic number$\sqrt[3]2$does not repeat. Continued Fractions periodicity and convolution on this site has a proof. 7 Of course there is a nicer proof! In fact, it's almost obvious if one thinks about geometric interpretation of continued fraction: consider the line$y=\alpha x$; then the best approximation (i.e. approximation that minimizes$|\alpha q-p|=q|\alpha-\frac{p}{q}|\$) is the point of integer lattice nearest to this line; finally observe that convergents with ...

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