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Let the number sequence $a_n$ be recursively defined by $a_0 = 1$ and

$a_n= a_{n-1}.(\frac{n}{n+1})^n$ if $n ≥ 0$

Does the series $\sum_{n=0}^{\infty}a_n$ diverge or converge? (Please note that it's the convergence of the series that I am trying to determine, not the sequence)

How should I handle the recursive formula of this problem? I did tried using the common convergence tests but none seemed to work. Can someone please give me a hint? Thank you all!

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2 Answers 2

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We have $(1+x)^n = 1 + xn + \ldots$, thus $(\frac{n+1}n)^n = (1+\frac1n)^n = 1 + n \frac1n + ... \ge 1 + n \frac1n = 2$ and $a_{n+1} = \frac{a_n}{(1+n^{-1})^n} \le \frac{a_n}2$. Thus $a_{n+1} \le \frac{a_n}2 \le \frac{a_{n-1}}4 \le \ldots \le \frac{a_1}{2^n}$. Hence $\sum_{n \ge 1} a_n < \infty$.

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  • $\begingroup$ Can you please explain how did you come to "∑n≥1 an<∞" through an+1≤an/2≤an−1/4≤…≤a1/2n ? $\endgroup$
    – Hung Nguyen
    Commented Oct 22, 2021 at 20:25
  • $\begingroup$ To what I understand you did prove that the sequence a(n) is decreasing and bounded. But how is that related to the series? Thank you $\endgroup$
    – Hung Nguyen
    Commented Oct 22, 2021 at 20:27
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    $\begingroup$ @HùngNguyễn here $a_n > 0$ and used the next fact: if $0 \le a_n \le b_n$ and $\sum_{n \ge 1} b_n < \infty$ then $\sum_{n \ge 1} a_n < \infty$. Here $b_n = \frac{a_1}{2^n} = 2^{-n}$ and hence $\sum_{n \ge 1} b_n < \infty$. $\endgroup$
    – Botnakov N.
    Commented Oct 22, 2021 at 20:53
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    $\begingroup$ Ah I think I got it now, thank you so much! $\endgroup$
    – Hung Nguyen
    Commented Oct 23, 2021 at 10:00
  • $\begingroup$ @ Hùng Nguyễn, you're welcome! $\endgroup$
    – Botnakov N.
    Commented Oct 23, 2021 at 11:39
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Hint: Take the log $b_n = \log a_n$.

DO NOT READ BELOW IF YOU ONLY WANT THE HINT.

$$b_n = b_{n-1} + n \ln \left (1 - \frac 1 {n+1}\right) = b_{n-1} -1 + \mathcal O\left(\frac 1 {n}\right)$$ Sum as a telescopic series: $b_n = b_0-n + \mathcal O(\ln n)$

Thus $a_n = \mathcal O(ne^{-n})$ and the series converges.

Edit for @Botnakov N. who was interested in a solution that doesn't use the estimate for the Harmonic series:

The same calculation shows $b_n = b_{n-1} + n \ln \left (1 - \frac 1 {n+1}\right) = b_{n-1} -1 + o(1)$

This entails that $b_n=b_0-n+o(n)$ and $a_n=\mathcal O(e^{-\frac n 2})$.

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  • $\begingroup$ Your solution imply that we know that $\sum_{1 \le k \le n} \frac{1}n = O(\ln n)$ and may easily work with big-O notation, but it looks like the author imply more simple technique. $\endgroup$
    – Botnakov N.
    Commented Oct 22, 2021 at 21:03
  • $\begingroup$ Knowing that the harmonic series is O(ln n) is not necessary in my solution, I added it to make it more interesting to the author. But yes, I am assuming knowledge of big-O notation. $\endgroup$
    – Stefan Lafon
    Commented Oct 22, 2021 at 23:23
  • $\begingroup$ @ Stefan Lafon, if it's not necessary then how will you get the asymptotic for $b_n$ from recurrent formula without it? $\endgroup$
    – Botnakov N.
    Commented Oct 22, 2021 at 23:36
  • $\begingroup$ The $\mathcal O(\frac 1 n)$ can be replaced by $o(1)$ and you still get an estimate that establishes the convergence. Would like me to spell it out for you? You seem so interested in that solution. $\endgroup$
    – Stefan Lafon
    Commented Oct 24, 2021 at 0:51
  • $\begingroup$ @ Stefan Lafon, thank you for personal addition, but the sum of $n$ times $o(1)$ is not $o(n)$ (counterexample: $a^{(i)}(n) = \frac{\sqrt{i}}{n}$ is $o(1)$ for any fixed $i$), so it's more precise to put $c_n = \ln(1-\frac{1}{n+1})+1 = o(1)$ and use the fact that $\sum_{1 \le k \le n} c_k = o(n)$ if $c_n \to 0$. Your solution works, but it looks like the author of problem (see comments) imply more simple technique then the fact about $\sum_{1 \le k \le n} c_k$. That's what I said and maybe you don't agree. But people may have different opinions :) $\endgroup$
    – Botnakov N.
    Commented Oct 24, 2021 at 9:10

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