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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).

Wheepy
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    Can you split it in two sums, by partial fraction decomposition? – Jens Wagemaker Apr 30 '18 at 19:25
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    Try finding the partial sum formula, or see the telescoping behavior. – Robbie Apr 30 '18 at 19:26
  • @DietrichBurde Thanks for pointing it out. I'd searched all over here, but it's difficult to find a specific series through the search form. I've typed it in different forms, to no avail. Any recommendations for searching this kind of thing in the future? I hate submitting duplicates. – Wheepy Apr 30 '18 at 19:27
  • How come you have studied math up to the gamma function, but you dont know about telescoping series ? – Rene Schipperus Apr 30 '18 at 19:28
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    @ReneSchipperus Erm, judgmental much? – Clement C. Apr 30 '18 at 19:29
  • @ClementC. Not a bit of it, just think its incongruous, and I am curious why. – Rene Schipperus Apr 30 '18 at 19:31
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    @ReneSchipperus https://stackoverflow.blog/2018/04/26/stack-overflow-isnt-very-welcoming-its-time-for-that-to-change/?cb=1 – Robbie Apr 30 '18 at 19:32
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    @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. – Wheepy Apr 30 '18 at 19:35
  • @RobbieVanDerzee I have seen that before, and I found it quite disgusting. This is the path to censorship, you need to be careful with that stuff cause it will burn you in ways you cant imagine. – Rene Schipperus Apr 30 '18 at 19:35
  • @Wheepy Hey stop attacking me. I just wanted to know where you are coming from. – Rene Schipperus Apr 30 '18 at 19:37
  • @ReneSchipperus I'm not attacking you. I used the conditional ("someone could interpret your comment"). I hoped you'd interpret me literally as you seemed to hope I'd done with your comment. It really was a heads up for the future. I know this forum, so it wasn't a problem for me. – Wheepy Apr 30 '18 at 19:46
  • Sorry folks, didn't mean to make a fuss. All's well. =) – Wheepy Apr 30 '18 at 19:53
  • @ReneSchipperus I just meant to be nice. – Robbie Apr 30 '18 at 20:07
  • @RobbieVanDerzee Yeah, well, I am not particularly pleased with this whole thing. Making a bit of banter is now not allowed. Not good. – Rene Schipperus Apr 30 '18 at 20:23

3 Answers3

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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 $N\to\infty$.

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

Clement C.
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    (Very often, when dealing with these sort of series where you expect nice simplifications, partial fraction expansion is a good way to go.) – Clement C. Apr 30 '18 at 19:25
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    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. – Wheepy Apr 30 '18 at 19:41
  • @Wheepy Sure, no worries. – Clement C. Apr 30 '18 at 19:43
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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}

Dispersion
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    Do you really think that OP knows digamma function and Euler constant? – Przemysław Scherwentke Apr 30 '18 at 19:46
  • I was just posting another solution in case he did, or somebody else was interested. – Dispersion Apr 30 '18 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. – Wheepy Apr 30 '18 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. – The Integrator Apr 30 '18 at 20:01
  • @The Integrator Yes, this is one of the reason why I posted this answer. – Dispersion Apr 30 '18 at 20:12
  • @TheIntegrator agreed. – Wheepy May 02 '18 at 14:10