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


  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.