# Find $\sum_{n=1}^{\infty} \frac{1}{n(n+2)}$

I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum. $$\begin{equation} \sum_{n=1}^{\infty} \frac{1}{n(n+2)} \end{equation}$$ First I set $$\frac{1}{n(n+2)}= \frac{A}{n}+\frac{B}{n+2}$$ and solve to get $$A=\frac{1}{2}$$ and $$B=-\frac{1}{2}$$. So I'm finding $$\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{2n+4})$$ and proceed with the partial sum expansion. At the end I am left with $$S_N=\frac{1}{2}+\frac{1}{4}-\frac{1}{2N+4}$$ by cancelling terms in between and then $$\lim_{N \to \infty}S_N=\frac{3}{4}$$.

Am I doing it correctly? Thank you for your time.

• yep – tilper Jan 3 at 14:04
• Thanks @tilper. – Allorja Jan 3 at 14:06
• Check again on your partial sums. I think you missed another negative term (though it tends to zero). – JavaMan Jan 3 at 14:07
• Yeah I know I am missing a negative term because it cancels with a positive one in the following second pair of parentheses but I can always cancel while tending to infinity and it gets reallyyyy small. Thank you though. – Allorja Jan 3 at 14:10

$$\sum^{\infty}_{n=1}\frac{1}{n(n+2)} = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\frac{1}{n}-\frac{1}{n+2}\bigg]$$
$$= \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\bigg(\frac{1}{n}-\frac{1}{n+1}\bigg)+\bigg(\frac{1}{n+1}-\frac{1}{n+2}\bigg)\bigg]$$
$$= \frac{1}{2}\bigg[\frac{1}{1}+\frac{1}{2}\bigg] = \frac{3}{4}$$
In case you're interested, you can also compute this series using a certain representation of the digamma function $$\psi(z+1)=-\gamma+\sum_{n\ge 1} \frac{z}{n(n+z)}$$. We are then looking for $$\frac{\psi(3)+\gamma}{2}$$. Using the following integral representation of the digamma function, $$\psi(s+1)=-\gamma+\int_0^1 \frac{1-x^s}{1-x}\,dx$$ we find that $$\psi(3)=-\gamma+\frac{3}{2}$$. Your sum is then $$\sum_{n\ge 1} \frac{1}{n(n+2)}=\frac{3}{4}$$