Show that
$$1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)} = \frac{\pi^2}{6}$$
Proof
By Partial Fraction Decomposition
$$\frac{1}{n^{2} \left(n + 1\right)}=\frac{-1}{n}+\frac{1}{n^{2}}+\frac{1}{n + 1}$$
Summing the series
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)} &= \sum_{n=1}^{\infty} \left( \frac{1}{n^2} – \frac{1}{n} + \frac{1}{n+1} \right) \\
&= \overbrace{\sum_{n=1}^{\infty} \frac{1}{n^2}}^{\text{Basel}} + \overbrace{\sum_{n=1}^{\infty} \left(-\frac{1}{n} + \frac{1}{n+1}\right)}^{\text{Telescoping}}
\end{align*}
Basel Series
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$
Telescoping Series
$$\sum_{n=1}^{\infty} \left(-\frac{1}{n} + \frac{1}{n+1}\right) = -1$$
Thus,
\begin{align*}
1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)} &= 1 + \overbrace{\sum_{n=1}^{\infty} \frac{1}{n^2}}^{\text{Basel}} + \overbrace{\sum_{n=1}^{\infty} \left(-\frac{1}{n} + \frac{1}{n+1}\right)}^{\text{Telescoping}} \\
&= 1 + \frac{\pi^2}{6} -1 \\
&= \frac{\pi^2}{6}
\end{align*}
$\blacksquare$
Question
I'm curious if that's a valid rigorous proof?
Best Answer
As Anne pointed out in the comments, all you need to use is the following theorem:
Now just put $a_n =\frac{1}{n^2}$ and $b_n = -\frac{1}{n} + \frac{1}{n+1}.$