題目

5. (15%) Find the sum of the series or show that it diverges.

\sum_{n=1}^{\infty}\frac{1}{n(n+3)}

解答

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先作部分分式分解：

\frac{1}{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right).

因此前 $N$ 項和為

\begin{align*} \sum_{n=1}^{N}\frac{1}{n(n+3)} =&\,\frac13\sum_{n=1}^{N}\left(\frac1n-\frac1{n+3}\right)\\[4mm] =&\,\frac13\left(1+\frac12+\frac13-\frac1{N+1}-\frac1{N+2}-\frac1{N+3}\right). \end{align*}

令 $N\to\infty$ ，最後三項趨近於 $0$ ，所以原級數收斂，且

\sum_{n=1}^{\infty}\frac{1}{n(n+3)} =\frac13\left(1+\frac12+\frac13\right) =\boxed{\frac{11}{18}}.