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The summation $\sum_{k=1}^{\infty} \frac{\phi^{-2k}}{k^2}=\text{Li}_2(1/\phi^2)=\frac{\pi^2}{15}-\ln^2 \phi$ [duplicate]
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Computing $\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du=\frac{\pi^2}{20}$, $\zeta(2)=\frac53 \sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}}{2^{4n}(2n+1)^2}$
I evaluated the required ...
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