Evaluate $\sum_{n=1}^{\infty} \left( \arctan \frac{4n - 1}{2} - \arctan \frac{4n - 3}{2} \right)$ 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!
 A: $\newcommand{\+}{^{\dagger}}
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$\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}
A: Let $a_n$ be the $n^{th}$ term of the series at hand. We have
$$\begin{align}
a_n = & \tan^{-1}\frac{4n-1}{2} - \tan^{-1}\frac{4n-3}{2}\\
= & \tan^{-1}\left(\frac{\frac{4n-1}{2}-\frac{4n-3}{2}}{1 + \frac{4n-1}{2}\frac{4n-3}{2}}\right)
= \tan^{-1}\left(\frac{1}{(2n-1)^2 + \frac34}\right)\\
\end{align}$$
Notice $\;\tan^{-1}(x) = \Im\log(1 + ix)\;$ for real $x$, we can rewrite $a_n$ as
$$a_n = \Im\left\{\log\left( 1 + \frac{i}{(2n-1)^2 + \frac34}\right)\right\}
= \Im\left\{\log\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\}
$$
Compare this with the factors in the 
infinite product expansion
of $\cosh x$:
$$\cosh x = \prod_{n=1}^{\infty}\left(1 + \frac{4x^2}{(2n-1)^2\pi^2}\right)$$
We find$\color{blue}{^{[1]}}$
$$\begin{align}
\sum_{n=1}^\infty a_n 
= &\Im\left\{\log\cosh\left(\frac{\pi}{2}\sqrt{\frac34+i}\right)\right\}
= \Im\left\{\log\cosh\left[\frac{\pi}{2}\left(1 + \frac{i}{2}\right)\right]\right\}\\
= & \Im\left\{\log\left[\cosh\frac{\pi}{2} \cos\frac{\pi}{4} + \sinh\frac{\pi}{2}\sin\frac{\pi}{4}i\right]\right\}
= \Im\left\{\log\left[ 1 + \tanh\frac{\pi}{2} i\right]\right\}\\
= & \tan^{-1}\left[\tanh\left(\frac{\pi}{2}\right)\right]
\end{align}$$
Notes


*

*$\color{blue}{[1]}$ Given any two complex numbers $u$ and $v$, $\log(uv)$ need not equal to $\log u + \log u$ in general. Instead, we have
$$\log(uv) = \log u + \log v + i2\pi N$$
for some integer $N$. So in principle,
$$\begin{align}
a_n &= \Im\left\{\log\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\}\\
\implies \sum_{n=1}^\infty a_n
&= \Im\left\{\log\prod_{n=1}^\infty\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\}
+ 2\pi N
\end{align}$$
for some integer $N$ only. However, $a_n$ is small enough and the sum falls within the range $(-\frac{\pi}{2},\frac{\pi}{2})$, the $N$ here is actually zero. The naive looking replacement:
$$\sum_{n=1}^\infty a_n \quad\longrightarrow\quad
\Im\left\{\log\cosh\left(\frac{\pi}{2}\sqrt{\frac34+i}\right)\right\}$$
does work.

