TITLE: Evaluate $\sum_{n=1}^{\infty} \left( \arctan \frac{4n - 1}{2} - \arctan \frac{4n - 3}{2} \right)$
QUESTION [11 upvotes]: I got stuck on the following series:
$$ \sum_{n=1}^{\infty} \left( \arctan \frac{4n - 1}{2} - \arctan \frac{4n - 3}{2} \right). $$
I can't seem to make an approach because there's $-3$ not $+3$. Please help!

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{n = 1}^{\infty}\bracks{%
     \arctan\pars{4n - 1 \over 2} - \arctan\pars{4n - 3 \over 2}}:\ {\large ?}}$

\begin{align}&\color{#c00000}{\sum_{n = 1}^{\infty}\bracks{%
\arctan\pars{4n - 1 \over 2} - \arctan\pars{4n - 3 \over 2}}}
=\sum_{n = 0}^{\infty}
\int_{\pars{4n + 1}/2}^{\pars{4n + 3}/2}{\dd x \over x^{2} + 1}
\end{align}

With $\ds{x \equiv {4n + 1 \over 2} + \xi}$:
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}\bracks{%
\arctan\pars{4n - 1 \over 2} - \arctan\pars{4n - 3 \over 2}}}
=\sum_{n = 0}^{\infty}
\int_{0}^{1}{\dd\xi \over \bracks{\pars{4n + 1}/2 + \xi}^{2} + 1}
\\[5mm] = &\
\int_{0}^{1}\sum_{n = 0}^{\infty}
{1 \over \pars{2n + 1/2 + \xi + \ic}\pars{2n + 1/2 + \xi - \ic}}\,\dd\xi
\\[5mm] = &\
{1 \over 4}\int_{0}^{1}\sum_{n = 0}^{\infty}
{1 \over \pars{n + 1/4 + \xi/2 + \ic/2}\pars{n + 1/4 + \xi/2 - \ic/2}}\,\dd\xi
\\[5mm] = &\
{1 \over 4}\int_{0}^{1}
{\Psi\pars{1/4 + \xi/2 + \ic/2} - \Psi\pars{1/4 + \xi/2 - \ic/2} \over \ic}\,\dd\xi
\\[5mm] = &\
\half\,\Im\int_{0}^{1}\Psi\pars{1/4 + \xi/2 + \ic/2}\,\dd\xi
\\[5mm] = &\
\Im\ln\pars{\Gamma\pars{3/4 + \ic/2} \over \Gamma\pars{1/4 + \ic/2}} =
\Im\ln\pars{%
\Gamma\pars{{3 \over 4} + {\ic \over 2}}\Gamma\pars{{1 \over 4} - {\ic \over 2}}}
\qquad\qquad\qquad\quad\pars{1}
\\[5mm] = &\
\Im\ln\pars{\pi\root{2} \over \cosh\pars{\pi/2} - \ic\sinh\pars{\pi/2}} =
\arctan\pars{\tanh\pars{\pi \over 2}}
\qquad\qquad\qquad\qquad\quad\pars{2}
\end{align}

In $\pars{1}$ and $\pars{2}$ we used identity ${\bf\mbox{6.1.32}}$ of
$\large\mbox{this table}$.
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}
\bracks{%
\arctan\pars{4n - 1 \over 2} - \arctan\pars{4n - 3 \over 2}}}
\\[5mm]  = &\
\bbox[10px,border:1px groove navy]{\arctan\pars{\tanh\pars{\pi \over 2}}} \approx 0.7422
\end{align}