Evaluate the limit of $\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$

We have that by Stolz-Cesaro

$$\frac{a_n}{b_n}=\frac{\sum_{k=1}^{n}\frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$

$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=\frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=\frac{n+2}{2n+1}\left(1+\frac{1}{n+1}\right)^{n} \to \frac e 2$$

Hint: $$\frac{(k+1)^k}{k^{k-1}} = k \left(1+\frac{1}{k}\right)^k$$ and $$ \left(1+\frac{1}{k}\right)^k = \exp\left(k \ln\left(1+\frac{1}{k}\right)\right) = \exp\left(1 + O(1/k)\right) = e + O(1/k)$$ Now, what can you say about $$\sum_{k=1}^n k (e + O(1/k))$$ ?