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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)

cmarangu
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    Please don't use such harsh language. I suppose one reason your question could have been downvoted is that you did not provide any attempts, or examples of other solutions that could have narrowed your question down, and helped others guide their efforts. For example, showing your attempts so a new solution doesn't coincide with one of your ways. – Saketh Malyala Nov 14 '19 at 05:03
  • EDIT: I would like to give others a chance to attempt the problem before I give my solution though. _good idea i'll do that_ – cmarangu Nov 14 '19 at 05:16
  • Is the answer $2e^{-2}$? – Szeto Nov 14 '19 at 05:33
  • Szeto yes, and I am glad I waited to se if anyone else can solve it – cmarangu Nov 14 '19 at 05:45
  • If you would like me to share my solution firsth then u tell me – cmarangu Nov 14 '19 at 05:46
  • The answer cannot be $2e^{-2} $ as this number is $< 1$. Numerically, the sequence of value of a truncated expression of the sum seems to converge to $2\sqrt{e} \sim 3.297442541400256$ – achille hui Nov 14 '19 at 05:56
  • achille hui thats what he meant – cmarangu Nov 14 '19 at 05:57
  • achille hui i actually got $2{e^{-2}}$ when solving it, what I had meant was $2{e^{1/2}}$ – cmarangu Nov 14 '19 at 06:03
  • @Szeto im curious as to how u solved it within 11 minutes. Have u seen this problem befoe? Or are u just a legend. – cmarangu Nov 14 '19 at 06:12

2 Answers2

3

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 :)

Szeto
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  • mindblowing. so was the mistake to call $\frac{1}{\left(-1\right)!}=0$? because $\lim_{x\to{-1}}{\frac{1}{\Gamma{\left(x+1\right)}}}=0$ – cmarangu Nov 15 '19 at 15:46
  • @cmarangu The usage of $(-1)!$ is not what makes you proof not rigorous. $(-1)!$ can be avoided by noticing that the sum can be reindexed to start at $i=1$ as the summand is zero when $i=0$. I say that your proof is not quite rigorous because you did not show that the manipulations for finite fractions can be extended to infinite tower fractions. – Szeto Nov 15 '19 at 16:04
  • ok. how do i do that though? is it by proving that tower fractions are just recursively defined fractions? Ahh I see. First u make a $V_n$ which is definded in terms of $V_{n-1}$ then u just use this proven formula to get the same series that al of us solvers of this problem got, and the rest is "trivial" – cmarangu Nov 15 '19 at 19:02
  • @cmarangu You’ve got it! – Szeto Nov 15 '19 at 23:46
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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 $\frac{1}{-1!}=0$)

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

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

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

cmarangu
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    Aside form the use of $(-1)!$, your answer looks fine. I will recommend you change some of the steps to $$\cdots + \sum_{i=0}^\infty \frac{2i}{2^i i!} = \cdots + \sum_{i=1}^\infty\frac{2i}{2^i i!} = \cdots +\sum_{i=1}^\infty \frac{1}{2^{i-1}(i-1)!} = \cdots $$ – achille hui Nov 14 '19 at 06:38
  • alright thanks for the praise also apparently u knew(?) that $\lim_{x\to{-1}}{\frac{1}{\Gamma{\left(x+1\right)}}}=0$ so calling 1/(-1)!=0 is almost ok. – cmarangu Nov 15 '19 at 15:54
  • $1/(-1)! = 0$ is almost okay but it is something one don't want to explain why it works... – achille hui Nov 15 '19 at 15:58
  • the <15 chars comment was just invisible spaces btw – cmarangu Nov 15 '19 at 16:02
  • very informative article https://en.wikipedia.org/wiki/Zero-width_space – cmarangu Nov 15 '19 at 16:03