求
\[ \lim_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}} \]
Solution:
\[  \sum_{k=1}^{n}\left(\frac{k}{n}\right)^{n}=\sum_{k=0}^{n-1}{\left(\frac{n-k}{n}\right)^{n}}=\sum_{k=0}^{n-1}{\left(1-\frac{k}{n}\right)^{n}} \]
\[ \left(1-\frac{k}{n}\right)^{n}=e^{n\ln(1-\frac{k}{n})}\leq e^{-k} \]
Thus.
\[ \varlimsup_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}}\leq\sum_{k=0}^{\infty}{e^{-k}}=\frac{e}{e-1} \]

for any $N\leq n-1$

\[  \sum_{k=0}^{N}{\left(\frac{n-k}{n}\right)^{n}}\leq \sum_{k=0}^{n-1}{\left(\frac{n-k}{n}\right)^{n}}\]

\[ \sum_{k=0}^{N}{\left(\frac{n-k}{n}\right)^{n}}\sim \sum_{k=0}^{N}e^{-k}=\frac{e^{-1}(1-e^{N+1})}{1-e^{-1}}\]

hence

\[ \varliminf_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}}\geq \frac{e^{-1}(1-e^{N+1})}{1-e^{-1}}\]

hold for any $N$,let $N\to \infty$,we have

\[ \lim_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}} =\frac{e}{e-1} \]
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事实上，这个问题可以加边成

\[ \lim_{n\to\infty}n\cdot\left(\frac{e}{e-1}-\sum_{i=1}^{n}\left(\frac{i}{n}\right)^{n}\right)=\frac{e(e+1)}{2(e-1)^3}\]
证明留给读者。