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数検1級 (8) 無限級数

Last updated at Posted at 2019-05-08

問題

次の級数の和を求めなさい。ただし正の整数 $n$ に対して $n!$ は $n$ の階乗を表します。

$$ \sum_{n=1}^{\infty} \frac{n^2}{n!}$$

知識

$e^x$ のマクローリン展開:

\mathrm{e}^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots

したがって、$x=1$ を代入することで、

\mathrm{e} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots = \sum_{n=0}^{\infty} \frac{1}{n!}1

解答

既知の級数 $\sum_{n=0}^{\infty} \frac{1}{n!} = \mathrm{e}$ に寄せるため、次のように変形する。

\begin{align}
\sum_{n=1}^{\infty} \frac{n^2}{n!} 
&= \sum_{n=1}^{\infty} \frac{n^2-n+n}{n!} \\
&= \sum_{n=1}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} \\
&= \sum_{n=2}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} && \text{(第1項は $n=1$ でゼロ)} \\
&= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(n-1)!} \\
&= \sum_{m=0}^{\infty} \frac{1}{m!} + \sum_{k=0}^{\infty} \frac{1}{k!} && \text{($m=n-2, k=n-1$ と定義)}\\
&= \mathrm{e} + \mathrm{e} \\
&= 2 \mathrm{e}.
\end{align}

感想

  • 無限級数は一般的な方針がないから怖い
  • テイラー展開でちょうどいいものを探す
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