(0,1]等距离取n个点的n次方和的极限

hbghlyj

math.stackexchange.com/questions/164074/how-to-evaluate-lim-limi ... -limits-k-1n-fracknn
$$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^n}{n^n}=\frac{e}{e-1}$$


An asymptotic expansion can be obtained as below. More terms can be included by using more terms in the expansions of $\exp$ and $\log$.
$$
\begin{aligned}
\sum_{k=0}^n\frac{k^n}{n^n}
&=\sum_{k=0}^n\left(1-\frac{k}{n}\right)^n\\
&=\sum_{k=0}^n\exp\left(n\log\left(1-\frac{k}{n}\right)\right)\\
&=\sum_{k=0}^{\sqrt{n}}\exp\left(n\log\left(1-\frac{k}{n}\right)\right)+O\left(ne^{-\sqrt{n}}\right)\\
&=\sum_{k=0}^{\sqrt{n}}\exp\left(-k-\frac{1}{2n}k^2+O\left(\frac{k^3}{n^2}\right)\right)+O\left(ne^{-\sqrt{n}}\right)\\
&=\sum_{k=0}^{\sqrt{n}}e^{-k}\exp\left(-\frac{1}{2n}k^2+O\left(\frac{k^3}{n^2}\right)\right)+O\left(ne^{-\sqrt{n}}\right)\\
&=\sum_{k=0}^{\sqrt{n}}e^{-k}\left(1-\frac{1}{2n}k^2+O\left(\frac{k^ 4}{n^2}\right)\right)+O\left(ne^{-\sqrt{n}}\right)\\
&=\sum_{k=0}^{\sqrt{n}}e^{-k}-\frac{1}{2n}\sum_{k=0}^{\sqrt{n}}k^2e^{-k}+O\left(\frac{1}{n^2}\right)\\
&=\frac{e}{e-1}-\frac{1}{2n}\frac{e(e+1)}{(e-1)^3}+O\left(\frac{1}{n^2}\right)
\end{aligned}
$$
Several steps use
$$
\sum_{k=n}^\infty e^{-k}k^m=O(e^{-n}n^m)
$$
which decays faster than any power of $n$.

Czhang271828

建议加入一致收敛章节的习题.

记 $u(k,n)=\left\{\begin{align*}
&0&&k>n,\\
&\left(1-\dfrac{k-1}{n}\right)^n&&k\leq n.
\end{align*}\right.$.

对给定的 $k=k_0$, $u(k_0,n)$ 关于 $n$ 单调非减, 且 $\sup_n|u(k_0,n)|=\dfrac{1}{e^{k_0-1}}$. 根据 Weierstrass 判别法, 函数项级数 $u_k(n):=u_k(n)$ 一致收敛.

据连续性定理有

$$
\begin{align*}
\lim_{n\to\infty}\sum_{k=1}^n u_k(n)=&\lim_{n\to\infty}
\sum_{k=1}^\infty u_k(n)\\
=&\sum_{k=1}^\infty
\lim_{n\to\infty} u_k(n)\\
=&\sum_{k=1}^\infty\dfrac{1}{e^{k-1}}\\
=&\dfrac{e}{e-1}.
\end{align*}
$$