<s>[INST] 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!
 [/INST] 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.

</s>