<s>[INST] 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.
 [/INST] $$\begin{aligned}\sum_{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}$$
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