Show that $1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)} = \frac{\pi^2}{6}$

Question

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?

Answer 1

As Anne pointed out in the comments, all you need to use is the following theorem:

If $\ \displaystyle\sum_{n=1}^{\infty } a_n = A,\quad \sum_{n=1}^{\infty } b_n = B,\ $ then $\ \displaystyle\sum_{n=1}^{\infty } (a_n+b_n) = A+B.$

Now just put $a_n =\frac{1}{n^2}$ and $b_n = -\frac{1}{n} + \frac{1}{n+1}.$