Calculate $\lim_{n\to \infty} \frac{\frac{2}{1}+\frac{3^2}{2}+\frac{4^3}{3^2}+...+\frac{(n+1)^n}{n^{n-1}}}{n^2}$ Calculate 
$$\lim_{n\to \infty} \frac{\displaystyle \frac{2}{1}+\frac{3^2}{2}+\frac{4^3}{3^2}+...+\frac{(n+1)^n}{n^{n-1}}}{n^2}$$
I have messed around with this task for quite a while now, but I haven't succeeded to find the solution yet.
Help is appreciated!
 A: We want to compute:
$$ \color{red}{L}=\lim_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n}k\cdot\left(1+\frac{1}{k}\right)^k.$$
The sequence given by $a_k=\left(1+\frac{1}{k}\right)^k $ is converging to $e$, hence
$$ b_k = \frac{k a_k + (k-1) a_{k-1} + \ldots + 2 a_2 + a_1}{k+(k-1)+\ldots+2+1} $$
is converging to $e$, too, by Cesàro. Since $k+(k-1)+\ldots+1 = \frac{k(k+1)}{2}$, it follows that:
$$ L = \color{red}{\frac{e}{2}}.$$
A: Combining $$\frac{1}{1-1/n}>1+\frac1n \ \ \ n>1 $$ with the fact that for all positive $n$ we have $$ \left(1+\frac1n\right)^n<e<\left(1+\frac{1}{n}\right)^{n+1}, $$ and after rewriting your sequence $a_n$ as $\frac{1}{n^2}\sum_{k=1}^n k(1+1/k)^k$, we find the general term $s_k$ of the sum satisfies $$k\left(e-\frac ek\right)<s_k<ke$$. Thus $$\frac{e}{2} \leftarrow \frac{e}{2}\frac{n(n+1)}{n^2} - \frac{e}{n} =\frac{e}{n^2} \sum_{k=1}^n(k-1)<a_n< \frac{e}{n^2} \sum_{k=1}^n k=\frac e2 \frac{n(n+1)}{n^2} \to \frac e2,$$ and by the squeeze theorem we conclude $\lim\limits_{n\to\infty} a_n=\displaystyle\frac e2$. 
