Tricky Limit when u use Riemann Sum and Cesaro

$\displaystyle \lim_{n\rightarrow \infty} \left[ \left(\frac{1}{n}\right)^{n}+\left(\frac{2}{n}\right)^{n}+\left(\frac{3}{n}\right)^{n}+\cdots +\left(\frac{n}{n}\right)^{n} \right] =\displaystyle \lim_{n\rightarrow\infty} \sum_{k=1}^{n} \left(\frac{k}{n} \right)^{n}$ $=\displaystyle \lim_{n\rightarrow\infty} \sum_{k=0}^{n-1} \left(1 - \frac{k}{n} \right)^{n}$ Use Formula $\displaystyle \lim_{x\rightarrow \infty} \left( 1+\frac{1}{x} \right)^{x}=e$ we have $\displaystyle \lim_{n\rightarrow\infty} \sum_{k=0}^{n} \left(1 - \frac{k}{n} \right)^{n} = \displaystyle \lim_{n\rightarrow\infty} \sum_{k=0}^{n} \left( \lim_{n\rightarrow\infty} \left(1 + \frac{1}{\frac{-1}{k} n} \right)^{\frac{-1}{k} n} \right)^{-k}$ $\displaystyle =\lim_{n\rightarrow\infty} \sum_{k=0}^{n-1} e^{-k}$ $\displaystyle =\sum_{k=0}^{\infty} e^{-k}$ Use infinite geometric sequence we get $\displaystyle =\sum_{k=0}^{\infty} e^{-k}=1+\frac{1}{e}+\frac{1}{e^{2}}+\frac{1}{e^{3}}+\cdots = \frac{1}{\displaystyle 1-\frac{1}{e}}=\boxed{\displaystyle \frac{e}{e-1}}\approx 1.5819...$