My real analysis textbook shows this result but gives no proof for it, it would be great if someone could show me why the following result holds.
Show
$$\sum_{n=1}^{\infty} \frac{H_n}{n(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n^{2}}$$
where $H_n = \sum_{k=1}^{n} 1/k$.
On
$$S=\sum_{n=1}^{\infty} H_n \left ( \frac{1}{n}-\frac{1}{1+n} \right)$$ $H_n=\int_{0}^{1} \frac{1-x^n}{1-x} dx.$ So $$ S=\sum_{n=1}^{\infty} \int_{0}^{1} \int_{0}^{1} \frac{1-x^n}{1-x} (t^{n-1} -t^{n})~dx~dt=\int_{0}^{1}\int_{0}^{1}\frac{1-t}{1-x} \sum_{n=1}^{\infty} (t^{n-1}-(tx)^{n}t^{-1})$$ $$S=\int_{0}^{1} \int_{0}^{1}\frac{1-t}{1-x} \frac{1-x)}{(1-x)(1-xt)} dt dx=\int_{0}^{1} \int_{0}^{1} \frac{1}{1-xt} dt dx =\sum_{n=1}^{\infty}\int_{0}^{1} \int_{0}^{1} x^nt^n dx dt.$$ $$\implies S=\sum_{n=1}^{\infty} \frac{1}{n^2}$$
Using $\sum_{n\ge1}\sum_{k=1}^na_{nk}=\sum_{k=1}^n\sum_{n\ge k}a_{nk}$, the sum is$$\sum_{n\ge1}\left(\frac1n-\frac{1}{n+1}\right)\sum_{k=1}^n\frac1k=\sum_{k\ge1}\frac1k\sum_{n\ge k}\left(\frac1n-\frac{1}{n+1}\right)=\sum_{k\ge1}\frac1k\frac1k.$$