# Evalaute $\sum_{k=1}^\infty \arctan \left(\frac{1}{1+3k+k^2}\right)$

Evalaute : $$\sum_{k=1}^\infty \arctan \left( \frac{1}{1+3k+k^2} \right)$$

This problem is supposed to be solved using the formula: $$\arctan(a)-\arctan(b)=\arctan \left(\frac{a-b}{1+ab} \right)$$ How ever i can't two such $$a$$ and $$b$$ that solve the problem .

I've experimented with $$a=k+3$$ and $$b=k$$ but it yields : $$\frac{3}{1+3k+k^2}$$ instead of $$\frac{1}{1+3k+k^2}$$

there's also $$a=k+2$$ and $$b=k+1$$ which give : $$\frac{1}{k^2+3k+2}$$ which doesn't solve the problem neither .

I've had many failed attempts and searched all over the net but couldn't find any thread on this problem.

• Are you sure that a closed form expression exists for this sum ? Commented Oct 11, 2023 at 14:23
• I am not actually , I only found this exercise online after trying for so many hours now , i couldn't tell if i was just unabe to solve it or if it dind't have a closed form so that's why i made this thread Commented Oct 11, 2023 at 14:30
• Mathematica says the sum is equivalent to the definite integral,$$\frac{\pi}2\int_0^1\left(\frac{\tan\left(\frac\pi2\sqrt{5-4ix}\right)}{\sqrt{5-4ix}}+\frac{\tan\left(\frac\pi2\sqrt{5+4ix}\right)}{\sqrt{5+4ix}}\right)\,dx$$ Commented Oct 11, 2023 at 15:52
• It appears interesting that $$\arctan\frac i2=\sum_{k=1}^\infty \arctan\left(\frac i{k^2+3k+1}\right)$$ so that the $\operatorname{arctanh}$ series returns $\operatorname{arctanh}\frac 12$. Commented Oct 11, 2023 at 16:03
• Using $(3.16)$ from Kunle Adegoke's paper about sums of arctangents. If you replace your $\arctan$ by $\operatorname{arctanh}$ you'll get $\operatorname{arctanh}\frac 12$ Commented Oct 11, 2023 at 16:31

using complex form we have $$\tan^{-1}x=\Im (\ln(1+xi))$$ then $$\sum_{k=1}^\infty \tan^{-1}\left(\frac{1}{k^2+3k+1} \right)=\Im\left(\sum_{k=1}^\infty \ln\left(1+\frac{i}{k^2+3k+1} \right)\right)$$ So $$\sum_{k=1}^\infty \tan^{-1}\left(\frac{1}{k^2+3k+1} \right)=\Im \left(\ln \prod_{k=1}^{\infty} \left(1+\frac{i}{k^2+3k+1} \right) \right)$$ to calculate the product we have $$\prod_{k=1}^{\infty} \left(1+\frac{i}{k^2+3k+1} \right)=\prod_{k=1}^{\infty} \frac{\left(k+\frac{3}{2}+\sqrt{\frac{5}{4}-i} \right)\left(k+\frac{3}{2}-\sqrt{\frac{5}{4}-i} \right)}{\left(k+\frac{3}{2}+\frac{\sqrt{5}}{2} \right)\left(k+\frac{3}{2}-\frac{\sqrt{5}}{2} \right)}$$ using the formula $$\prod_{k=1}^{\infty} \prod_{p=1}^{n} \frac{k+a_p}{k+b_p}=\prod_{p=1}^{n} \frac{\Gamma(b_p+1)}{\Gamma(a_p+1)} , \text{ where } \sum_{k=1}^n (a_k-b_k)=0$$ So $$\prod_{k=1}^{\infty} \left(1+\frac{i}{k^2+3k+1} \right)= \frac{\Gamma\left(1+\frac{3}{2}+\frac{\sqrt{5}}{2} \right)\Gamma\left(1+\frac{3}{2}-\frac{\sqrt{5}}{2} \right)}{\Gamma\left(1+\frac{3}{2}+\sqrt{\frac{5}{4}-i} \right)\Gamma\left(1+\frac{3}{2}-\sqrt{\frac{5}{4}-i} \right)}$$ we can show easily that $$\Gamma\left(1+\frac{3}{2}+\frac{\sqrt{5}}{2} \right)\Gamma\left(1+\frac{3}{2}-\frac{\sqrt{5}}{2} \right)=-\pi \sec \left(\frac{\sqrt{5} \pi}{2} \right)$$ and $$\Gamma\left(1+\frac{3}{2}+\sqrt{\frac{5}{4}-i} \right)\Gamma\left(1+\frac{3}{2}-\sqrt{\frac{5}{4}-i} \right)=-2\pi \sec \left(\frac{\pi}{2} \sqrt{5-4i} \right)$$ So $$\Im \left(\ln \frac{-\pi \sec \left(\frac{\sqrt{5} \pi}{2} \right)}{-2\pi \sec \left(\frac{\pi}{2} \sqrt{5-4i} \right)} \right)=\arg\left(-\cos \left(\frac{\pi}{2} \sqrt{5-4i} \right) \right)$$ by long computing we can get $$\arg\left(-\cos \left(\frac{\pi}{2} \sqrt{5-4i} \right) \right)=\tan^{-1} \left(\text{tanh}\left(\frac{\pi\sqrt{-5+\sqrt{41}}}{2\sqrt{2}} \right)\tan\left(\frac{\pi\sqrt{5+\sqrt{41}}}{2\sqrt{2}} \right) \right)$$ finally $$\sum_{k=1}^\infty \tan^{-1}\left(\frac{1}{k^2+3k+1} \right)=\tan^{-1} \left(\text{tanh}\left(\frac{\pi\sqrt{-5+\sqrt{41}}}{2\sqrt{2}} \right)\tan\left(\frac{\pi\sqrt{5+\sqrt{41}}}{2\sqrt{2}} \right) \right)$$