一道求极限题目

求极限: $\displaystyle\mathop {\lim }\limits_{n \to \infty } n\left[ {\frac{e}{{e - 1}} - \sum\limits_{k = 1}^n {{{\left( {\frac{k}{n}} \right)}^n}} } \right]$
已邀请:
$\lim\limits_{n\to \infty}n\bigl(\begin{smallmatrix}\cfrac{e}{e-1}-\sum_{k=1}^{n}{(\cfrac{k}{n})}^{n}\end{smallmatrix}\bigr)
=\lim\limits_{n\to \infty}n\bigl(\begin{smallmatrix}\sum_{k=0}^{\infty}{e}^{-k}-\sum_{k=0}^{n-1}{(1-\cfrac{k}{n})}^{n}\end{smallmatrix}\bigr)
\\=\lim\limits_{n\to \infty}n\bigl(\begin{smallmatrix}\sum_{k=0}^{n-1}{e}^{-k}-{(1-\cfrac{k}{n})}^{n}+\sum_{k=n}^{\infty}{e}^{-k}\end{smallmatrix}\bigr)
=\lim\limits_{n\to \infty}n\bigl(\begin{smallmatrix}\sum_{k=0}^{n-1}{e}^{-k}(1-{(1-\cfrac{k}{n})}^{n}{e}^{k})\end{smallmatrix}\bigr)$$
\\=\lim\limits_{n\to \infty}n\bigl(\begin{smallmatrix}\sum_{k=0}^{n-1}{e}^{-k}(-ln{(1-\cfrac{k}{n})}^{n}-k+O({(-ln{(1-\cfrac{k}{n})}^{n}-k)}^{2})\end{smallmatrix}\bigr)$$
=\lim\limits_{n\to \infty}n\bigl(\begin{smallmatrix}\sum_{k=0}^{n-1}{e}^{-k}(n(\cfrac{k}{n}+\cfrac{{k}^{2}}{2{n}^{2}}+O(\cfrac{{k}^{3}}{3{n}^{3}}))-k+O({(-ln{(1-\cfrac{k}{n})}^{n}-k)}^{2})\end{smallmatrix}\bigr)$$
\\=\lim\limits_{n\to \infty}n\bigl(\begin{smallmatrix}\sum_{k=0}^{n-1}{e}^{-k}(\cfrac{{k}^{2}}{2n}+O(\cfrac{{k}^{3}}{{n}^{2}}))+O({(\cfrac{{k}^{2}}{2n})}^{2})\end{smallmatrix}\bigr)$$
=\lim\limits_{n\to \infty}\sum_{k=0}^{n-1}{e}^{-k}\cfrac{{k}^{2}}{2}+\cfrac{O(\sum_{k=0}^{n-1}{e}^{-k}{k}^{3})}{n}+\cfrac{O(\sum_{k=0}^{n-1}{e}^{-k}{k}^{4})}{4n}
\\=\lim\limits_{n\to \infty}\sum_{k=0}^{n-1}{e}^{-k}\cfrac{{k}^{2}}{2}
=\sum_{k=0}^{\infty}{e}^{-k}\cfrac{{k}^{2}}{2}
=S=\sum_{k=1}^{\infty}{e}^{-k+1}\cfrac{{(k-1)}^{2}}{2}
\\=eS-\sum_{k=1}^{\infty}{e}^{-k+1}\cfrac{2k-1}{2}
=\cfrac{1}{2e-2}\sum_{k=1}^{\infty}{e}^{-k+1}(2k-1)
\\=\cfrac{1}{2e-2}\sum_{k=0}^{\infty}{e}^{-k}(2k+1)
=\cfrac{1}{2e-2}+{e}^{-1}S+\cfrac{1}{2e-2}\sum_{k=1}^{\infty}2{e}^{-k}
\\=\cfrac{1}{1-{e}^{-1}}\cfrac{1}{2e-2}(1+\sum_{k=1}^{\infty}2{e}^{-k})
=\cfrac{{e}^{-1}({e}^{-1}+1)}{2{(1-{e}^{-1})}^{3}}


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