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?

  • $\begingroup$ 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 $1$. :) $\endgroup$ – Hagen von Eitzen Jun 20 '13 at 22:18
  • $\begingroup$ (I added some parentheses to help clarify the meaning of the expression. Feel free to remove them if you disagree with their inclusion.) $\endgroup$ – Andrés E. Caicedo Jun 21 '13 at 4:40
  • $\begingroup$ Hmm, I would expect, that the list in the exponents terminates at some final value. Otherwise we get, for instance $1/0$ somewhere... $\endgroup$ – Gottfried Helms Jun 21 '13 at 5:17
  • 1
    $\begingroup$ 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 $n \gt N \in \Bbb N$. $\endgroup$ – Daniel Margolis Jun 21 '13 at 5:26

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:

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

  • $\begingroup$ [Leaves a cheerful comment because he cannot upvote] $\endgroup$ – Pedro Tamaroff Jun 20 '13 at 23:03
  • $\begingroup$ @Lord_Farin brilliant (+1) $\endgroup$ – user 1357113 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 $a=\lim_n a_{n+1}=(\lim_n b_n)^{\lim_n a_n}=1^a=1$.

  • $\begingroup$ 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. $\endgroup$ – Andrés E. Caicedo Jun 21 '13 at 2:07
  • $\begingroup$ 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. $\endgroup$ – Daniel Margolis Jun 21 '13 at 3:22
  • 1
    $\begingroup$ Daniel, I am sorry, but what you wrote makes no sense. $\endgroup$ – Andrés E. Caicedo Jun 21 '13 at 4:36
  • $\begingroup$ I don't mean to be a hard-ass, just things I noticed. $\endgroup$ – Daniel Margolis Jun 21 '13 at 4:37
  • 1
    $\begingroup$ 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. $\endgroup$ – Andrés E. Caicedo Jun 21 '13 at 4:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.