Can anyone help me solve these sequence limits?

By the Stolz-Cesro Theorem , one has \begin{eqnarray} &&\lim_{n\to\infty} \frac{1}{n^2} \left(2+ \frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}}\right)\\ &=&\lim_{n\to\infty} \frac{\sum_{k=1}^n\frac{(k+1)^k}{k^{k-1}}}{n^2}\\ &=&\lim_{n\to\infty} \frac{\sum_{k=1}^{n+1}\frac{(k+1)^k}{k^{k-1}}-\sum_{k=1}^n\frac{(k+1)^k}{k^{k-1}}}{(n+1)^{2}-n^2}\\ &=&\lim_{n\to\infty} \frac{\frac{(n+2)^{n+1}}{(n+1)^{n}}}{2n+1}\\ &=&\lim_{n\to\infty} \frac{n+2}{2n+1}\frac{(n+2)^{n}}{(n+1)^{n}}\\ &=&\lim_{n\to\infty} \frac{n+2}{2n+1}\left(1+\frac{1}{n+1}\right)^n\\ &=&\frac e2. \end{eqnarray}

For the first $$\frac {(n+1)^n}{n^{n-1}}=(n+1)\left(1+\frac 1n\right)^{n-1}=\frac{n+1}{1+\frac 1n}\left(1+\frac 1n\right)^{n}\to (n+1)e$$

For the second, if you erase all the $+1$s you decrease the denominator and increase the fraction. But it makes the fraction $$\frac{\sqrt{(n-1)!}}{\sqrt{n!}}=\frac 1{\sqrt n} \to 0$$