Series involving arctan

Evaluate the series:

$$\sum_{n=1}^{\infty} \arctan \frac{1}{n^{2}} $$

Solution



We begin by the identity:

$$\log(1+ix)=\frac{1}{2} \log(1+x^2)+i\tan^{-1}x  \tag{1}$$

Summing $(1)$ we get that:

$$\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{k^2}\right)=\Im\left[\log\left(\prod_{k=1}^\infty\left(1-\frac{\alpha^2}{k^2}\right)\right)\right] \tag{2} $$

where $\displaystyle  \alpha =\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$ and $a^2 = -i$.

We are also invoking the famous identity:

$$\prod_{k=1}^\infty\left(1-\frac{\alpha^2}{k^2}\right)=\frac{\sin(\pi\alpha)}{\pi
\alpha}  \tag{3}$$

However we note that:

$$\begin{aligned}
\frac{\sin(x+iy)}{x+iy}
&=\frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{x+iy}\\
&=\frac{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}{x^2+y^2}\\
&=i\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x^2+y^2}
\end{aligned} \tag{4}$$

Combining $(2), \; (3), \; (4)$ along with the identity $\displaystyle \tan(\pi/4-x)=\frac{1-\tan(x)}{1+\tan(x)}$ we get that:

$$\begin{aligned}
\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{k^2}\right)
&=\Im\left(\log\left(\frac{\sin(\pi\alpha)}{\pi\alpha}\right)\right)\\
&=\tan^{-1}\left(\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}\right)\\
&=\tan^{-1}\left(\frac{\tan(\frac{\pi}{\sqrt{2}})-\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})+\tanh(\frac{\pi}{\sqrt{2}})}\right)\\
&=\frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right)
\end{aligned}$$

which is the desired result.