# Show that the series converges to 1 [duplicate]

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).

• Can you split it in two sums, by partial fraction decomposition? Commented Apr 30, 2018 at 19:25
• Try finding the partial sum formula, or see the telescoping behavior. Commented Apr 30, 2018 at 19:26
• @ReneSchipperus Erm, judgmental much? Commented Apr 30, 2018 at 19:29
• @ReneSchipperus stackoverflow.blog/2018/04/26/… Commented Apr 30, 2018 at 19:32
• @ReneSchipperus It just happens that it's been a very long time since I've seen series convergence and I hadn't thought about telescoping series. I've always thought of this forum as a friendly environment to ask questions, even if they seem kind of dumb. I've included the context in which I've found it because I know this is encouraged here so that who's answering knows how much of a complete answer the person who's asking needs, and in fact the tips here were sufficient for me. Someone could interpret your comment as shaming and be discouraged to continue asking here or to provide context. Commented Apr 30, 2018 at 19:35

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 $N\to\infty$.

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

• (Very often, when dealing with these sort of series where you expect nice simplifications, partial fraction expansion is a good way to go.) Commented Apr 30, 2018 at 19:25
• Although you were faster, I think I must choose the other answer for completeness (for the sake of future reference, in case this question isn't closed as a duplicate). I hope this is ok with you. Commented Apr 30, 2018 at 19:41
• @Wheepy Sure, no worries. Commented Apr 30, 2018 at 19:43

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}

• Do you really think that OP knows digamma function and Euler constant? Commented Apr 30, 2018 at 19:46
• I was just posting another solution in case he did, or somebody else was interested. Commented Apr 30, 2018 at 19:51
• I agree with @PrzemysławScherwentke - I do not know it yet, and it seems a little backwards to prove the convergence using such an advanced result (although it's a creative solution). In fact, I've found this series in Havil's book on the Gamma function, in the part he's introducing the digamma function. So I really needed the convergence before the expansion form of digamma. Commented Apr 30, 2018 at 19:52
• Even tho OP probably would much prefer the telescoping sum. The digamma function usage might be of some help to future readers. This would also help tie in different aspects of math imo. Commented Apr 30, 2018 at 20:01
• @The Integrator Yes, this is one of the reason why I posted this answer. Commented Apr 30, 2018 at 20:12