compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$ Knowing that $1 - \frac 12 + \frac 13 - \cdots = \ln 2$ and $1 - \frac 13 + \frac 15 - \cdots = \frac{\pi}{4}$, compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$. I programmed and found that $\frac{\ln 2}{4}+{\frac{\pi}{4}}/2$ is precise to the eighth digit.
 A: $$\begin{aligned}\sum_{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}$$
A: $$
{\cal F}\left(x\right)
\equiv
\sum_{n = 0}^{\infty}{x^{4n + 2} \over \left(4n +1\right)\left(4n + 2\right)}
$$
\begin{align}
{\cal F}''\left(x\right)
&\equiv
\sum_{n = 0}^{\infty}x^{4n}
=
{1 \over 1 - x^{4}}
=
{1/2 \over 1 - x^{2}} + {1/2 \over 1 + x^{2}}
=
{1 \over 4}\sum_{\sigma = \pm}{1 \over 1 + \sigma\,x} + {1/2 \over 1 + x^{2}}
\\[3mm]
{\cal F}'\left(x\right)
&\equiv
{1 \over 4}\sum_{\sigma = \pm}\sigma\,\ln\left(1 + \sigma\,x\right)
+
{1 \over 2}\arctan\left(x\right)
\\[3mm]
{\cal F}\left(x\right)
&\equiv
{1 \over 4}\sum_{\sigma = \pm}\sigma\left\lbrack%
x\ln\left(1 + \sigma\,x\right)
-
\int_{0}^{x}x'\,{\sigma \over 1 + \sigma\,x'}\,{\rm d}x'
\right\rbrack
+
{1 \over 2}\,x\,\arctan\left(x\right)
-
{1 \over 2}\int_{0}^{x}{x' \over x'^{2} + 1}\,{\rm d}x'
\\[3mm]&=
{1 \over 4}\sum_{\sigma = \pm}\sigma\left\lbrack%
x\ln\left(1 + \sigma\,x\right)
-
x
+
\sigma\ln\left(1 + \sigma\,x\right)
\right\rbrack
+
{1 \over 2}\,x\arctan\left(x\right)
-
{1 \over 4}\,\ln\left(1 + x^{2}\right)
\\[5mm]&
\end{align}
$$
{\cal F}\left(x\right)
=
{1 \over 4}\left\lbrack%
\left(-x + 1\right)\ln\left(1 - x\right) + \left(x + 1\right)\ln\left(1 + x\right)
\right\rbrack
+
{1 \over 2}\,x\arctan\left(x\right)
-
{1 \over 4}\,\ln\left(1 + x^{2}\right)
$$
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\lim_{x \to 1^{-}}{\cal F}\left(x\right)
=
\sum_{n = 0}^{\infty}{1 \over \left(4n +1\right)\left(4n + 2\right)}
=
{1 \over 8}\,\pi + {1 \over 4}\,\ln\left(2\right)\quad}
\\ \\ \hline
\end{array}
$$
A: $$
\begin{align}
\log(2)&=\sum_{n=0}^\infty\frac1{4n+1}-\frac1{4n+2}+\frac1{4n+3}-\frac1{4n+4}\\
\frac12\log(2)&=\sum_{n=0}^\infty\hphantom{-\frac1{4n+2}}\frac1{4n+2}\hphantom{\;+\frac1{4n+3}}-\frac1{4n+4}\\
\frac\pi4&=\sum_{n=0}^\infty\frac1{4n+1}\hphantom{-\frac1{4n+2}}-\frac1{4n+3}\\
\frac12\left(\log(2)-\frac12\log(2)+\frac\pi4\right)&=\sum_{n=0}^\infty\frac1{4n+1}-\frac1{4n+2}
\end{align}
$$
