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

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 $e.$
