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I'm trying to calculate $\sum_{k=1}^\infty\frac{1}{(4k-1)(4k+4)}$ using telescopic sums. I've already proved this equality: $\sum\frac{1}{(4k-1)(4k+4)}=\frac{1}{5}\big(\sum\frac{1}{4k-1}-\frac{1}{4k+4}\big)$. The problem is I can't cancel the terms of this sum.

I need help

Thanks

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  • $\begingroup$ Is there a reason to think you can turn this into a telescoping sum? Seems difficult. You might be able to turn it into a sum of logarithms. $\endgroup$ – Thomas Andrews Mar 31 '15 at 2:32
  • $\begingroup$ @ThomasAndrews The book I'm using suggested to use telescopic sums. $\endgroup$ – user42912 Mar 31 '15 at 2:33
  • $\begingroup$ According to Wolfram, the solution is $\frac{1}{40}(2-\pi + \log(64))$. Given that, I have upvoted you because I am curious how an answer like that is going to arise from a telescoping sum. $\endgroup$ – JessicaK Mar 31 '15 at 2:42
  • $\begingroup$ the sum can be expressed as $\frac{-\psi (-1/4)-\gamma }{20}$ where $\psi $ is the digamma function and $\gamma $ is the Euler constant $\endgroup$ – Lozenges Mar 31 '15 at 3:04
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Hint: $S=\displaystyle \lim_{n\to \infty} S_n=\displaystyle \lim_{n\to \infty} \displaystyle \sum_{k=1}^n \left(\dfrac{1}{4k-1} - \dfrac{1}{4k+4}\right) = \displaystyle \lim_{n\to \infty} \displaystyle \int_{0}^1 \left(\displaystyle \sum_{k=1}^n\left(x^{4k-2} - x^{4k+3}\right)\right)dx= \displaystyle \int_{0}^1 \displaystyle \lim_{n\to \infty} \displaystyle \sum_{k=1}^n \left(x^{4k-2} - x^{4k+3}\right)dx$. Can you compute the integrand, and integrate it.

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  • $\begingroup$ Aren't we looping in such a case ? $\endgroup$ – Claude Leibovici Mar 31 '15 at 3:43
  • $\begingroup$ I've done this before and even asked a question like it before and was sure it can be done this way, and unlikely loop. $\endgroup$ – DeepSea Mar 31 '15 at 3:45

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