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

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

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

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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

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

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 $e.$

Since terms of the sum in the denominator approach $\frac{1}{e}$ and there are $n$ of them, the limit of the sequence is $e$.