问题及解答

(1)

\[
\sum_{n=1}^{\infty}\frac{8}{n(n+2)}
\]

解.

\[
\begin{split}
S_n&=\sum_{i=1}^{n}\frac{8}{i(i+2)}\\
&=\sum_{i=1}^{n}4\biggl(\frac{1}{i}-\frac{1}{i+2}\biggr)\\
&=4\biggl(\sum_{i=1}^{n}\frac{1}{i}-\sum_{i=1}^{n}\frac{1}{i+2}\biggr)\\
&=4\biggl((\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n})\\
&\quad\quad-(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2})\biggr)\\
&=4\biggl(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\biggr)
\end{split}
\]

于是

\[
\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}4\biggl(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\biggr)=6.
\]