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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?

luxerhia
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RiRi
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1 Answers1

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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 $\sum_{k=1}^{\infty}\frac{4k}{4k^{4}+1}=1$.

Alessio K
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  • I'm so so sorry, i just have 1 doubt, how did you make 1/(2$k^2$-2k+1) into 1? – RiRi Aug 16 '20 at 14:33
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    When you take a finite sum you'll see that all terms except the first and last term cancel, try it! – Alessio K Aug 16 '20 at 14:34
  • +1. You have a "good eye". Nice answer. – Felix Marin Aug 16 '20 at 14:46
  • +1. Is there a way to see that the latter sum is telescopic apart from summing some of the initial terms? – Bernkastel Aug 16 '20 at 14:53
  • I'm not sure if there is any faster way, but it's easy to see from the first few terms. Maybe see here https://math.stackexchange.com/questions/104918/how-to-analyze-convergence-and-sum-of-a-telescopic-series-i-cant-find-a-generi and https://en.wikipedia.org/wiki/Telescoping_series. – Alessio K Aug 16 '20 at 15:05