Zeenat Horn

2022-02-24

Prove $\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{k}\sum _{j=1}^{2k}\frac{{\left(-1\right)}^{j}}{j}=\frac{{\pi }^{2}}{48}+\frac{1}{4}{\mathrm{ln}}^{2}2$

Kathryn Duggan

Use
$\sum _{j=1}^{n}\frac{{\left(-1\right)}^{j-1}}{j}=\mathrm{ln}2+{\left(-1\right)}^{n-1}{\int }_{0}^{1}\frac{{x}^{n}}{1+x}dx$
we get
$\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{k}\sum _{j=1}^{2k}\frac{{\left(-1\right)}^{j}}{j}=\mathrm{ln}2\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{k}-\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{k}{\int }_{0}^{1}\frac{{x}^{2k}}{1+x}dx$
$={\mathrm{ln}}^{2}2+{\int }_{0}^{1}\frac{1}{1+x}\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-{x}^{2}\right)}^{k}}{k}dx={\mathrm{ln}}^{2}2-{\int }_{0}^{1}\frac{\mathrm{ln}\left(1+{x}^{2}\right)}{1+x}dx$
let$f\left(t\right)={\int }_{0}^{1}\frac{\mathrm{ln}\left(1+t{x}^{2}\right)}{1+x}dx$
then
${f}^{\prime }\left(t\right)={\int }_{0}^{1}\frac{{x}^{2}}{\left(1+x\right)\left(1+t{x}^{2}\right)}dx$
$=\frac{1}{t+1}{\int }_{0}^{1}\frac{x-1}{1+t{x}^{2}}dx+\frac{1}{t+1}{\int }_{0}^{1}\frac{dx}{x+1}$
$=\frac{1}{t+1}{\left[\frac{1}{2t}\mathrm{ln}\left(1+t{x}^{2}\right)-\frac{1}{\sqrt{t}}\mathrm{arctan}\left(\sqrt{t}x\right)+\mathrm{ln}\left(x+1\right)\right]}_{0}^{1}$

Tye Rhodes

Note that
$\sum _{j=1}^{2k}\frac{{\left(-1\right)}^{j}}{j}=\sum _{j=1}^{k}\frac{1}{2j}-\sum _{j=1}^{k}\frac{1}{2j-1}=\frac{1}{2}\sum _{j=1}^{k}\frac{1}{j}-\left(\sum _{j=1}^{2k}\frac{1}{j}-\sum _{j=1}^{k}\frac{1}{2j}\right)=\sum _{j=1}^{k}\frac{1}{j}-\frac{j=1}{2k}\frac{1}{j}$
$={H}_{k}-{H}_{2k}$
Then,
$\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{k}\sum _{j=1}^{2k}\frac{{\left(-1\right)}^{j}}{j}=\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{k}\left({H}_{k}-{H}_{2k}\right)$
$=\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{k}{H}_{k}-2\sum _{k=1}^{\mathrm{\infty }}\frac{{i}^{2k}}{2k}{H}_{2k}$
$=f\left(-1\right)-2R\left(f\left(i\right)\right)$ where $f\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{{x}^{k}}{k}{H}_{k}$
However,
$f\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}{x}^{k}{H}_{k}{\int }_{0}^{1}{t}^{k-1}dt={\int }_{0}^{1}\frac{1}{t}\sum _{k=1}^{\mathrm{\infty }}{\left(xt\right)}^{k}{H}_{k}dt={\int }_{0}^{1}\frac{1}{t}\left[-\frac{\mathrm{ln}\left(1-xt\right)}{1-xt}\right]dt$

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