Find a sum of $\sum_{n=1}^{\infty} \tan^{-1} (\frac{2}{n^2 + n + 1})$

Question

Find a sum of $$\sum_{n=1}^{\infty} \tan^{-1} \left(\frac{2}{n^2 + n + 1}\right)$$

Is it possible? I tried to make telescopic formula for such $\tan^{-1}$, but all my trials had no success.

P.S. I've tried to do it in the same way as for $\sum_{n=1}^{\infty} \tan^{-1} \left(\frac{1}{n^2 + n + 1}\right)$ and $\sum_{n=1}^{\infty} \tan^{-1} \left(\frac{2}{n^2 + n + 4}\right)$, so I tried to represent the fraction $\frac{2}{n^2 + n + 1}$ as $\frac{a - b}{1 + ab}$:

$$ \begin{cases} a - b = 2 \\ 1 + ab = 1 + n + n^2 \end{cases} $$ but this system has solutions: $$ \begin{cases} a = 1 - \sqrt{n^2 + n + 1} \\ b = - 1 - \sqrt{n^2 + n + 1} \end{cases}$$ $$ \begin{cases} a = \sqrt{n^2 + n + 1} + 1 \\ b = \sqrt{n^2 + n + 1} - 1 \end{cases}$$

and

$$\tan^{-1} \left(\frac{2}{n^2 + n + 1}\right) = \tan^{-1} \left(\sqrt{n^2 + n + 1} + 1\right) - \tan^{-1} \left(\sqrt{n^2 + n + 1} - 1\right)$$

This difference does not allow us to make sum $\sum_{n=1}^{\infty} \tan^{-1} \left(\frac{2}{n^2 + n + 1}\right)$ to be a telescopic sum as it was in cases $\sum_{n=1}^{\infty} \tan^{-1} \left(\frac{1}{n^2 + n + 1}\right)$ and $\sum_{n=1}^{\infty} \tan^{-1} \left(\frac{2}{n^2 + n + 4}\right)$

Would here be some other ways to find accurate sum of $\sum_{n=1}^{\infty} \tan^{-1} \left(\frac{2}{n^2 + n + 1}\right)$?

This might be the same idea: https://math.stackexchange.com/questions/193001/explicitly-finding-the-sum-of-arctan1-n2n1 — Gerry Myerson, Oct 10 '24 at 02:42
This question is similar to: $\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Gerry Myerson, Oct 10 '24 at 02:43
$$\ldots=\pi-\tan^{-1}(2)+\tan^{-1}\left[\tan\left(\pi\sqrt{\frac{\sqrt{73}-3}{8}}\right)\tanh\left(\pi\sqrt{\frac{\sqrt{73}+3}{8}}\right)\right].$$ — metamorphy, Oct 10 '24 at 04:42
@GerryMyerson thank you, I've edited my question and added steps, which explain why this series is different from you mentioned — Ivan Zhuk, Oct 10 '24 at 10:39

score 5 · Answer 1 · 2024-10-10T15:49:31.200

$$\sum_{n=1}^{\infty}\arctan\left(\frac{2}{n^2+n+1}\right)$$ $$=\sum_{n=1}^{\infty}\Im \log\left(1+\frac{2i}{n^2+n+1}\right)=\Im\log\prod_{n=1}^{\infty}\frac{(n^2+n+1+2i)}{(n^2+n+1)}$$ $$=\Im\log\prod_{n=1}^{\infty}\frac{(n+\frac{1-\sqrt{-3-8i}}{2})(n+\frac{1+\sqrt{-3-8i}}{2})}{(n+\frac{1-\sqrt{3}i}{2})(n+\frac{1+\sqrt{3}i}{2})}=\Im\log\frac{b_1b_2\Gamma(b_1)\Gamma(b_2)}{a_1a_2\Gamma(a_1)\Gamma(a_2)}$$ Where $a_1=\frac{1+\sqrt{-3-8i}}{2}$, $a_2=\frac{1-\sqrt{-3-8i}}{2}$,$b_1=\frac{1-\sqrt{3}i}{2}$ and $b_2=\frac{1+\sqrt{3}i}{2}$ Using Euler reflection formula gives $$=\Im \log\frac{\cos(\frac{\pi }{2}\sqrt{-3-8i})}{(1+2i)\cos(\frac{\sqrt{3}}{2}\pi i)}=\arg\left((1-2i)\cos\left(\frac{\pi }{2}\sqrt{-3-8i}\right)\right)$$

Note that $\sqrt{-3-8i}={\underbrace{\frac{\sqrt{\sqrt{73}-3}}{\sqrt{2}}}_{a}}{\underbrace{-i\frac{\sqrt{\sqrt{73}+3}}{\sqrt{2}}}_{-ib}}$

$$\sum_{n=1}^{\infty}\arctan\left(\frac{2}{n^2+n+1}\right)=\arg((1-2i)(\cos a\cosh b +i\sin a\sinh b))$$ By numerical exploitation we have to pick the $+\pi$ branch if we were to use $\arg(x+iy)=\arctan(y/x)$ for real $x,y$. $$=\pi-\arctan(2)+\arctan(\tan a\tanh b)$$ $$=\pi-\arctan(2)+\arctan\left(\tan \frac{\pi\sqrt{\sqrt{73}-3}}{2\sqrt{2}} \tanh \frac{\pi\sqrt{\sqrt{73}+3}}{2\sqrt{2}}\right)$$

Thank you. Does there exist the way to get an accurate sum using Real Analysis, not Complex Analysis? — Ivan Zhuk, Oct 10 '24 at 13:46
@IvanZhuk If you want an approximation, since we know the integral, we can use the simplest form of Euler-MacLaurin summation — Claude Leibovici, Jan 10 '25 at 15:42

Find a sum of $\sum_{n=1}^{\infty} \tan^{-1} (\frac{2}{n^2 + n + 1})$

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