Show that the series converges to 1

Question

I want to show the following:

$$ \sum_{r=1}^\infty \frac{1}{r(r+1)} = 1. $$

I've found this series as part of a calculation to prove a formula for the Gamma function. I know it converges to 1 because of the result of this calculation, but otherwise I wouldn't even know how to find this value, so any help is appreciated with how to find out the series converges to 1 and/or how to prove it converges to 1 (in case these steps are done separately).

Answer 1

When you see a factored polynomial in the denominator, think partial fractions. Consider the partial sums $$\sum_{r=1}^N\frac{1}{r(r+1)}=\sum_{r=1}^N\left(\frac{1}{r}-\frac{1}{r+1}\right)$$ this sum is telescoping, and then take $N\to\infty$.

Answer 2

Hints: $$ \frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1} $$ and "telescopic series".

Answer 3

You can also use a series expansion form of the digamma functions: $$\psi^{(0)}(s+1)=-\gamma+\sum_{n=1}^{\infty} \frac{s}{n(n+s)}$$ You are then looking for

\begin{align} \psi^{(0)}(2)+\gamma&=(1-\gamma)+\gamma \\ &=\boxed{1} \end{align}