级数 $\lim_{n \to \infty}$ $\sum_{n = 1}^{n}$ $\mathrm{e}^{\frac{i}{n}}$ 求和怎么计算？

一、题目

$$
I = \lim_{n \to \infty} \sum_{n = 1}^{n} \mathrm{e}^{\frac{i}{n}} = ?
$$

难度评级：

二、解析

$$
\begin{aligned}
I \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \sum_{n = 1}^{n} \mathrm{e}^{\frac{i}{n}} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \left( \mathrm{e}^{\frac{1}{n}} + \mathrm{e}^{\frac{2}{n}} + \cdots + \mathrm{e}^{\frac{n-2}{n}} + \mathrm{e}^{\frac{n-1}{n}} + \mathrm{e}^{\frac{n}{n}} \right) \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \left( \textcolor{yellow}{ \mathrm{e} + \mathrm{e}^{\frac{n-1}{n}} + \mathrm{e}^{\frac{n-2}{n}} + \cdots + \mathrm{e}^{\frac{2}{n}} + \mathrm{e}^{\frac{1}{n}} } \right)
\end{aligned}
$$

又因为：

$$
\begin{cases}
\mathrm{e}^{\textcolor{orangered}{ \frac{n-1}{n} }} = \mathrm{e}^{1 + \textcolor{orangered}{ \frac{n-1}{n} } – 1} = \mathrm{e} ^{1 + \frac{-1}{n}} = \mathrm{e} \cdot \mathrm{e}^{\frac{-1}{n}} = \textcolor{springgreen}{ \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{1} } \\
\quad \Downarrow \\
\mathrm{e}^{\frac{n-2}{n}} = \mathrm{e} \cdot \mathrm{e}^{\frac{-2}{n}} = \textcolor{springgreen}{ \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{2} } \\
\quad \Downarrow \\
\mathrm{e}^{\frac{n-3}{n}} = \mathrm{e} \cdot \mathrm{e}^{\frac{-3}{n}} = \textcolor{springgreen}{ \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{3} } \\
\quad \Downarrow \\
\quad \vdots \\
\quad \Downarrow \\
\mathrm{e}^{\frac{2}{n}} = \mathrm{e} \cdot \mathrm{e}^{\frac{2-n}{n}} = \textcolor{springgreen}{ \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{n-2} } \\
\quad \Downarrow \\
\mathrm{e}^{\frac{1}{n}} = \mathrm{e} \cdot \mathrm{e}^{\frac{1-n}{n}} = \textcolor{springgreen}{ \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{n-1} } \\
\end{cases}
$$

于是：

$$
\begin{aligned}
I \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \left( \textcolor{yellow}{ \mathrm{e} + \mathrm{e}^{\frac{n-1}{n}} + \mathrm{e}^{\frac{n-2}{n}} + \cdots + \mathrm{e}^{\frac{2}{n}} + \mathrm{e}^{\frac{1}{n}} } \right) \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \left[ \mathrm{e} \cdot 1 + \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{1} + \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{2} + \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{3} + \cdots \right. \\
– & \ \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{n-2} + \mathrm{e} \cdot (\mathrm{e}^{\frac{-1}{n}})^{n-1} \left]\right. \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \mathrm{e} \cdot \left[ \textcolor{pink}{ 1 + (\mathrm{e}^{\frac{-1}{n}})^{1} + (\mathrm{e}^{\frac{-1}{n}})^{2} + (\mathrm{e}^{\frac{-1}{n}})^{3} + \cdots + (\mathrm{e}^{\frac{-1}{n}})^{n-2} + (\mathrm{e}^{\frac{-1}{n}})^{n-1} } \right]
\end{aligned}
$$

由于 $\textcolor{pink}{1}$ $\textcolor{pink}{+}$ $\textcolor{pink}{(\mathrm{e}^{\frac{-1}{n}})^{1}}$ $\textcolor{pink}{+}$ $\textcolor{pink}{(\mathrm{e}^{\frac{-1}{n}})^{2}}$ $\textcolor{pink}{+}$ $\textcolor{pink}{(\mathrm{e}^{\frac{-1}{n}})^{3}}$ $\textcolor{pink}{+}$ $\textcolor{pink}{\cdots}$ $\textcolor{pink}{+}$ $\textcolor{pink}{(\mathrm{e}^{\frac{-1}{n}})^{n-2}}$ $\textcolor{pink}{+}$ $\textcolor{pink}{(\mathrm{e}^{\frac{-1}{n}})^{n-1}}$ 其实就是一个首项为 $1$, 公比为 $\mathrm{e}^{\frac{-1}{n}}$ 的等比数列的前 $n$ 项和（含有 “$\mathrm{e}^{\frac{-1}{n}}$” 的项有 $n-1$ 个，再加上一个首项 “$1$”, 所以共有 $n$ 项），即：

$$
\begin{aligned}
& \textcolor{pink}{1 + (\mathrm{e}^{\frac{-1}{n}})^{1} + (\mathrm{e}^{\frac{-1}{n}})^{2} + (\mathrm{e}^{\frac{-1}{n}})^{3} + \cdots + (\mathrm{e}^{\frac{-1}{n}})^{n-2} + (\mathrm{e}^{\frac{-1}{n}})^{n-1}} \\ \\
= & \ \frac{1 \cdot \left[ 1 – (\mathrm{e}^{\frac{-1}{n}})^{n} \right]}{1 – \mathrm{e}^{\frac{-1}{n}}} \\ \\
= & \ \textcolor{springgreen}{ \frac{\left[ 1 – (\mathrm{e}^{\frac{-1}{n}})^{n} \right]}{1 – \mathrm{e}^{\frac{-1}{n}}} }
\end{aligned}
$$

于是：

$$
\begin{aligned}
I \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \sum_{n = 1}^{n} \mathrm{e}^{\frac{i}{n}} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \mathrm{e} \cdot \left[ \textcolor{pink}{ 1 + (\mathrm{e}^{\frac{-1}{n}})^{1} + (\mathrm{e}^{\frac{-1}{n}})^{2} + (\mathrm{e}^{\frac{-1}{n}})^{3} + \cdots + (\mathrm{e}^{\frac{-1}{n}})^{n-2} + (\mathrm{e}^{\frac{-1}{n}})^{n-1} } \right] \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \mathrm{e} \cdot \frac{\left[ 1 – (\mathrm{e}^{\textcolor{magenta}{ \frac{-1}{n}} })^{\textcolor{magenta}{n}} \right]}{1 – \mathrm{e}^{\frac{-1}{n}}} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \frac{\mathrm{e} \left[ 1 – (\mathrm{e}^{\textcolor{magenta}{-1}}) \right]}{1 – \mathrm{e}^{\frac{-1}{n}}} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \frac{\mathrm{e} \left( 1 – \frac{1}{\mathrm{e}} \right)}{\frac{\mathrm{e}^{\frac{1}{n}} – 1}{\mathrm{e}^{\frac{1}{n}}}} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \frac{\mathrm{e} \left( 1 – \frac{1}{\mathrm{e}} \right)}{1} \cdot \frac{\mathrm{e}^{\frac{1}{n}}}{\mathrm{e}^{\frac{1}{n}} – 1} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \frac{\mathrm{e} – 1}{1} \cdot \frac{\mathrm{e}^{\frac{1}{n}}}{\mathrm{e}^{\frac{1}{n}} – 1} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \frac{\textcolor{yellow}{\mathrm{e}^{\frac{1}{n}}} \cdot (\mathrm{e} – 1)}{\textcolor{pink}{\mathrm{e}^{\frac{1}{n}} – 1}} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \frac{\textcolor{yellow}{1} \cdot (\mathrm{e} – 1)}{\textcolor{pink}{\frac{1}{n}}} \\ \\
= & \ \textcolor{gray}{ \lim_{n \to \infty} } \textcolor{springgreen}{\boldsymbol{ n \cdot (\mathrm{e} – 1) }}
\end{aligned}
$$

Tip

由于当 $n \rightarrow \infty$ 的时候，$\mathrm{e}^{\frac{1}{n}}(\mathrm{e} – 1)$ $\rightarrow$ $\mathrm{e} – 1$, 所以，对于 $\textcolor{gray}{ \lim_{n \to \infty} } \frac{\mathrm{e}^{\frac{1}{n}}(\mathrm{e} – 1)}{\mathrm{e}^{\frac{1}{n}} – 1}$ 我们不能使用洛必达法则做进一步的化简。

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