jmc

We can write $$\frac{1}{(n+1)H_n{H}_{n+1}}=\frac{\frac{1}{n+1}}{H_n{H}_{n+1}}=\frac{H_{n+1}-H_n}{H_n{H}_{n+1}}=\frac{1}{H_n}-\frac{1}{H_{n+1}}.$$Thus, $$\begin{array}{rl}\sum _{n=1}^{\infty }\frac{1}{(n+1)H_n{H}_{n+1}}& =\sum _{n=1}^{\infty }\left(\frac{1}{H_n}-\frac{1}{H_{n+1}}\right)\\ & =\left(\frac{1}{H_1}-\frac{1}{H_2}\right)+\left(\frac{1}{H_2}-\frac{1}{H_3}\right)+\left(\frac{1}{H_3}-\frac{1}{H_4}\right)+\cdots \\ & =\frac{1}{H_1}=\boxed{1}.\end{array}$$Note that this result depends on the fact that $H_n\to \infty$ as $n\to \infty .$ We can prove this as follows: $$\begin{array}{rl}\frac{1}{2}& \ge \frac{1}{2},\\ \frac{1}{3}+\frac{1}{4}& >\frac{1}{4}+\frac{1}{4}=\frac{1}{2},\\ \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}& >\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2},\end{array}$$and so on. Thus, $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots >1+\frac{1}{2}+\frac{1}{2}+\cdots ,$$which shows that $H_n\to \infty$ as $n\to \infty .$