We can simply use comparison test to know whether this series converges or diverges, obviously this one converges but how do we find the actual value after summation?

Can we use integration? I'm preparing for an exam and they are permitting 3 minutes to max 5 minutes per question, so how can i tackle questions like this that will help me find sum under 5 minutes?

  • 2
    $\begingroup$ partial fractions? $\endgroup$ Aug 16, 2020 at 14:14
  • $\begingroup$ Why do you think the sum converges? $\endgroup$
    – user619894
    Aug 16, 2020 at 14:14
  • $\begingroup$ wolframalpha.com/input/… According to WA, the sum is equal to 1. $\endgroup$
    – Crostul
    Aug 16, 2020 at 14:24
  • 1
    $\begingroup$ Note $4k^4+1= (2k^2+2k+1)(2k^2-2k+1)$ which is easily obtained by Sophia Germaine identity and rest is telescoping sum. He same problem is here. $\endgroup$
    – Naren
    Aug 16, 2020 at 14:58
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    $\begingroup$ @Naren that was really useful, thank you so much for showing the expansion!! $\endgroup$
    – RiRi
    Aug 16, 2020 at 16:53

1 Answer 1


We have $\frac{4k}{4k^{4}+1}=\frac{4k}{(4k^{4}+4k^{2}+1)-4k^{2}}=\frac{4k}{(2k^{2}+2k+1)(2k^{k}-2k+1)}=\frac{1}{2k^{2}-2k+1}-\frac{1}{2k^{2}+2k+1}$ and thus $\sum_{k=1}^{n}\frac{4k}{4k^{4}+1}=1-\frac{1}{2n^{2}+2n+1}$, so finally $\sum_{k=1}^{\infty}\frac{4k}{4k^{4}+1}=1$.

  • $\begingroup$ I'm so so sorry, i just have 1 doubt, how did you make 1/(2$k^2$-2k+1) into 1? $\endgroup$
    – RiRi
    Aug 16, 2020 at 14:33
  • 1
    $\begingroup$ When you take a finite sum you'll see that all terms except the first and last term cancel, try it! $\endgroup$
    – Alessio K
    Aug 16, 2020 at 14:34
  • $\begingroup$ +1. You have a "good eye". Nice answer. $\endgroup$ Aug 16, 2020 at 14:46
  • $\begingroup$ +1. Is there a way to see that the latter sum is telescopic apart from summing some of the initial terms? $\endgroup$
    – Bernkastel
    Aug 16, 2020 at 14:53
  • $\begingroup$ I'm not sure if there is any faster way, but it's easy to see from the first few terms. Maybe see here math.stackexchange.com/questions/104918/… and en.wikipedia.org/wiki/Telescoping_series. $\endgroup$
    – Alessio K
    Aug 16, 2020 at 15:05

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