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Does the sequence $$\displaystyle \frac{n}{\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^k}$$ converge?

Attempt. Since $\Big(\frac{k}{k+1}\Big)^k \rightarrow 1/e\neq 0$ and the terms are positive, the series $\sum\limits_{k=1}^{\infty}\Big(\frac{k}{k+1}\Big)^k$ diverges to $+\infty$. I find hard to determine if $n$ or the sum goes faster to $+\infty.$

Thanks in advance.

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  • $\begingroup$ If $u_n\sim v_n$, $v_n\geq 0$ and $\sum v_n$ diverges, then $\sum_{k=1}^n u_k\sim \sum_{k=1}^n v_k$, so the answer is positive. $\endgroup$
    – Gabriel Romon
    Sep 3, 2018 at 16:54

3 Answers 3

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Recall that by Stolz-Cesaro we have

$$\lim_{n\to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}=L \implies \lim_{n\to \infty} \frac{a_n}{b_n}=L$$

and in that case we have

$$\lim_{n\to \infty}\frac{n}{\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^k}=\lim_{n\to \infty}\frac{1}{\Big(\frac{n+1}{n+2}\Big)^{n+1}}=\lim_{n\to \infty}{\Big(\frac{n+2}{n+1}\Big)^{n+1}}=\lim_{n\to \infty}{\Big(1+\frac{1}{n+1}\Big)^{n+1}}\to e$$

To determine the rate of convergence and also as an alternative to solve the limit, we have that

$${\Big(\frac{k}{k+1}\Big)^k}=e^{k\log \Big(\frac{k}{k+1}\Big)}=e^{-k\log \Big(1+\frac{1}{k}\Big)}=e^{-k\Big(\frac{1}{k}-\frac{1}{2k^2}+O(k^{-3})\Big)}e^{-1+\frac{1}{2k}+O(k^{-2})}=\frac1e\left(1+\frac{1}{2k}+O(k^{-2})\right)$$

and therefore

$$\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^k=\frac1e\sum\limits_{k=1}^{n}\left(1+\frac{1}{2k}+O(k^{-2})\right)\sim\frac1e\left(n+\frac12\ln n\right)$$

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If $a_n\to L,$ then as is well known, $(a_1+\cdots + a_n)/n \to L.$ Since $[n/(n+1)]^n \to 1/e,$ we therefore have

$$\frac{\sum_{k=1}^{n}[k/(k+1)]^k}{n} \to \frac{1}{e}.$$

Taking reciprocals gives the limit of $e.$

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  • $\begingroup$ The well known fact is Stolz-Cesaro theorem or are you referring to some other result? Thanks $\endgroup$
    – user
    Sep 3, 2018 at 18:12
  • $\begingroup$ @gimusi Many readers will be familiar with the result I cited without knowing S-C. $\endgroup$
    – zhw.
    Sep 3, 2018 at 19:08
  • $\begingroup$ I'm not referring to the name of the theorem but to the theorem itself. I'm interested to know whether the well known fact is something different and simpler than that or it is the same theorem. In that case maybe it might be useful to give a reference to that result in order to do not let the false impression that this kind of "intuitive" reasoning always works. $\endgroup$
    – user
    Sep 3, 2018 at 19:30
  • $\begingroup$ @gimusi Intuitive reasoning? I cited a well-known result. Many readers will know it from a basic real analysis course. Baby Rudin has it as an exercise for example. It is implied by S-C, but is a simpler result. $\endgroup$
    – zhw.
    Sep 4, 2018 at 16:24
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Since terms of the sum in the denominator approach $\frac{1}{e}$ and there are $n$ of them, the limit of the sequence is $e$.

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