# Can we use integration to find $\sum_{k=1}^\infty \frac {4k}{4{k}^{4}+1}$?

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?

• partial fractions? Aug 16, 2020 at 14:14
• Why do you think the sum converges? Aug 16, 2020 at 14:14
• wolframalpha.com/input/… According to WA, the sum is equal to 1. Aug 16, 2020 at 14:24
• 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. Aug 16, 2020 at 14:58
• @Naren that was really useful, thank you so much for showing the expansion!!
– RiRi
Aug 16, 2020 at 16:53

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 $$\sum_{k=1}^{\infty}\frac{4k}{4k^{4}+1}=1$$.
• I'm so so sorry, i just have 1 doubt, how did you make 1/(2$k^2$-2k+1) into 1?