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What is the value of the following limit? $$\large \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{\frac{n}{n-1}^{\frac{n-1}{n-2}^{...}}}$$ In general what do limits of infinite decreasing numbers strung together in familiar ways approach?

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If only you had written $\left(\left(\frac{n+1}n^{\frac n{n-1}}\right)^{\frac{n-1}{n-2}}\right)^\cdots$. Then I'd know the limit is $1$. :) – Hagen von Eitzen Jun 20 '13 at 22:18
(I added some parentheses to help clarify the meaning of the expression. Feel free to remove them if you disagree with their inclusion.) – Andrés E. Caicedo Jun 21 '13 at 4:40
Hmm, I would expect, that the list in the exponents terminates at some final value. Otherwise we get, for instance $1/0$ somewhere... – Gottfried Helms Jun 21 '13 at 5:17
I made a comment on an answer that was deleted that offered to terminate the sequence when $n-k=0$. So it would just be ${...}^{n-k+1}$. k is, with respect to sequence limits, an n st $n \gt N \in \Bbb N$. – Daniel Margolis Jun 21 '13 at 5:26
up vote 22 down vote accepted

The sequence is given by:

$$a_1 = 2, a_n = \left(\frac{n+1}n\right)^{a_{n-1}}$$

Taking logs, we obtain:

$$\log a_n = a_{n-1} \log\left(1+\frac1n\right)$$

and it is easy to show that $\dfrac1{2n} \le \log\left(1+\frac1n\right) \le \dfrac1n$.

Now if we can show that $a_n$ is bounded, we are done by the Squeeze theorem (since then $\lim\limits_{n\to\infty} \log a_n = 0$, hence $\lim\limits_{n\to\infty} a_n = 1$).

Obviously, $a_n \ge 0$ for all $n$. We prove inductively that $a_n \le e$. The basis is trivial: $a_1 = 2 \le e$. Suppose $a_{n-1} \le e$. By the above estimate, $\log a_n \le \dfrac en \le 1$ for $n \ge 3$. It remains to show that $a_2 \le e$:

$$a_2 = \left(\dfrac32\right)^2 = \dfrac94 \le e$$

In conclusion:

$$\lim_{n\to\infty} a_n = \lim_{n \to \infty} {\large\frac{n+1}{n}^{\frac{n}{n-1}^{\frac{n-1}{n-2}^{...}}}} = 1$$

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[Leaves a cheerful comment because he cannot upvote] – Pedro Tamaroff Jun 20 '13 at 23:03
@Lord_Farin brilliant (+1) – user 1618033 Jun 21 '13 at 5:36

For $n\ge 1$, let $$a_n=\left(\frac{n+1}{n}\right)^{\frac{n}{n-1}^{\frac{n-1}{n-2}^{...}}}$$ where the tower stops when we reach $2/1=2$, so $a_1=2$ and $$a_{n+1}=\left(\frac{n+2}{n+1}\right)^{a_n}.$$

Note that each $a_n$ is (strictly) larger than $1$, and that the sequence is decreasing from $n=2$ on: First, $a_2=9/4>2>1.911>a_3$. Next, if $a_{n+1}<a_n$, then $$a_{n+2}=\left(\frac{n+3}{n+2}\right)^{a_{n+1}}<\left(\frac{n+2}{n+1}\right)^{a_{n+1}}<\left(\frac{n+2}{n+1}\right)^{a_n}=a_{n+1}. $$ Since the sequence is decreasing and bounded below by $1$, it follows that $\lim_n a_n=a$ exists, and satisfies $a\ge 1$.

It remains to argue that $a=1$. To see this, note that $a_{n+1}=b_n^{a_n}$, where $\displaystyle b_n=\frac{n+2}{n+1}$, so $a=\lim_n a_{n+1}=(\lim_n b_n)^{\lim_n a_n}=1^a=1$.

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Daniel, you seem to have misunderstood the argument. (Or maybe it is confusion with basic properties of sequences and convergence.) I will be away of the computer for a bit, but if you specify what lines seem confusing, I can attempt to point you to relevant results I may be using. – Andrés E. Caicedo Jun 21 '13 at 2:07
There can only be one limit of the sequence $a_n=a$, but your theory assumes that if for some $n \gt N, a=345.7$, then $a_{n+1}=1$ and $a=1$ automatically. But $a$ can only equal one value, a sequence can only converge once. By construction, I claim your proof is flawed by including the unknown in input to get the unknown in output. – Daniel Margolis Jun 21 '13 at 3:22
Daniel, I am sorry, but what you wrote makes no sense. – Andrés E. Caicedo Jun 21 '13 at 4:36
I don't mean to be a hard-ass, just things I noticed. – Daniel Margolis Jun 21 '13 at 4:37
I'm sorry, but it looks like you do not understand basic properties of sequences and limits. You have written four comments here, and three of them make no sense. As I said above, state specifically what line seems confusing. – Andrés E. Caicedo Jun 21 '13 at 4:43

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