$1+\frac{3+\frac{5+\frac{⋰}{6}}{4}}{2}$

Question

Suppose we have a recursively defined sequence $T_n$ such that $T_n=n+\frac{T_{n+2}}{n+1}$ and so $T_1$ looks like this:

$$1+\cfrac{3+\cfrac{5+\cfrac{7+\cfrac{9+\cfrac{⋰}{10}}{8}}{6}}{4}}{2}$$

How do I solve this?

(PostScriptum. I called it T for Tower because its similar to a Power-Tower)

Answer 1

My approach is slightly more rigorous than the OP's, as it is not generally justified to manipulate 'infinite tower fractions' as if it is finite.

However mine is much slower.

We have the recursion $$T_n=n+\frac{T_{n+2}}{n+1}$$ and the OP asks for $T_1$.

First, define $V_n=T_{2n+1}$. Then, $$V_n=2n+1+\frac{V_{n+1}}{2n+2}\implies V_{n}=-2n(2n-1)+2nV_{n-1}$$ and we want to find $V_0$.

Recall the extremely useful formula for first order recursion: $$f_n=\alpha_n+\beta_n f_{n-1}\implies f_n=f_0\prod^n_{i=1}\beta_i+\sum^{n-1}_{k=0}\alpha_{n-k}\prod^k_{j=1}\beta_{n-j+1}$$

Applying it gives $$\begin{align} V_n &=V_0\prod^n_{i=1}(2i)-\sum^{n-1}_{k=0}(2(n-k)-1)(2(n-k))\prod^k_{j=1}2(n-j+1) \\ &=V_02^n n!-\sum^{n}_{k=1}(2k-1)(2k)\prod^{n-k}_{j=1}2(n-j+1) \\ &=V_02^n n!-\sum^{n}_{k=1}(2k-1)(2k)\prod^{n}_{j=k+1}2j \\ &=V_02^n n!-\sum^{n}_{k=1}(2k-1)(2k)\cdot 2^{n-k}\frac{n!}{k!} \\ \frac{V_n}{2^nn!}&=V_0-\sum^{n}_{k=1}\frac{4k^2-2k}{2^kk!} \qquad{(\star)} \\ \end{align} $$

From the recurrence relation, it is apparent that $V_n=O(n)$. Therefore, taking limit $n\to\infty$ on both sides of $(\star)$, $$V_0-\sum^{\infty}_{k=1}\frac{4k^2-2k}{2^kk!}=0$$

We can apply known formulae to evaluate this sum very quickly, but I would like to demonstrate that closed form of this sum can be derived purely by elementary methods: $$\begin{align} \sum^{\infty}_{k=1}\frac{4k^2-2k}{2^kk!} &=\sum^{\infty}_{k=1}\frac{4k-2}{2^k(k-1)!} \\ &=\sum^{\infty}_{k=0}\frac{4k+2}{2^{k+1} k!} \\ &=2\sum^{\infty}_{k=0}\frac{k}{2^k k!}+\sum^{\infty}_{k=0}\frac{1}{2^k k!} \\ &=2\sum^{\infty}_{k=1}\frac{k}{2^k k!}+\sqrt e \\ &=2\sum^{\infty}_{k=1}\frac{1}{2^k (k-1)!}+\sqrt e \\ &=2\sum^{\infty}_{k=0}\frac{1}{2^{k+1} k!}+\sqrt e \\ &=\sum^{\infty}_{k=0}\frac{1}{2^{k} k!}+\sqrt e \\ V_0&=2\sqrt e \qquad{\blacksquare}\\ \end{align} $$

p.s. I have not seen this problem before.

p.p.s. We made the same careless mistake :)

Answer 2

My solution:

$\Large{1+\frac{3+\frac{5+\frac{⋰}{6}}{4}}{2}}$

$\Large{={1+\frac{3}{2}+\frac{\frac{5}{4}}{2}+\frac{\frac{\frac{7}{6}}{4}}{2}}⋯}$

$\Large{=\sum\limits_{i=0}^{\infty}{\frac{2i+1}{{2^i}\cdot{i}!}} }$

note that i is an iterator not an imaginary quantity

$\Large{=\sum\limits_{i=0}^{\infty}{\frac{1}{{2^i}\cdot{i}!}+\frac{2i}{{2^i}\cdot{i}!}} }$

$\Large{=\sum\limits_{i=0}^{\infty}{\frac{1}{{2^i}\cdot{i}!}+\frac{2i}{{2^i}\cdot{i}!}} }$

$\Large{=\sum\limits_{i=0}^{\infty}{\frac{1}{{2^i}\cdot{i}!}}+\sum\limits_{i=0}^{\infty}{\frac{2i}{{2^i}\cdot{i}!}} }$

$\Large{=\sum\limits_{i=0}^{\infty}{\frac{\left(\frac{1}{2}\right)^i}{{i}!}}+\sum\limits_{i=0}^{\infty}{\frac{1}{{2^{i-1}}\cdot\left(i-1\right)!}} }$

$\Large{=e^{1/2}+\sum\limits_{i=1}^{\infty}{\frac{1}{{2^{i-1}}\cdot\left(i-1\right)!}} }$

(1. because e^k sum formula from wikipedia 2. becuause $\frac{1}{-1!}=0$)

$\Large{=\sqrt{e}+\sum\limits_{i=0}^{\infty}{\frac{1}{{2^{i}}\cdot{i}!}} }$

$\LARGE{=2\sqrt{e}}$

$\LARGE{Q. E. D.}$