Evaluate $\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} - \frac{1}{2} \sum_ {m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1}$

Question

Let's declare $\mathcal{G}$ is constant of Catalanand the $\mathcal{H}_m-st$ mharmonic term. Let it be shown that:

$$\displaystyle{\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} -\frac{1}{2} \sum_ {m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1} = \frac{\pi \log 2}{8} - \frac{\mathcal {G}}{2}}$$

My try

Since $$\displaystyle{{H_{2m}} - \frac{1}{2}{H_m} = \left( {1 + \frac{1}{2} + \frac{1}{3} + .. + \ frac{1}{{2m}}} \right) - \frac{1}{2}\left( {1 + \frac{1}{2} + \frac{1}{3} + .. + \ frac{1}{m}} \right) = \sum\limits_{n = 1}^m {\frac{1}{{2n - 1}}} }$$ we have:

$$\displaystyle{S = \sum\limits_{m = 1}^\infty {\frac{{{{\left( { - 1} \right)}^m}}}{{2m + 1}}\left( {{H_{2m}} - \frac{1}{2}{H_m}} \right)} = \sum\limits_{m = 1}^\infty {\frac{{{{\left( { - 1 } \right)}^m}}}{{2m + 1}}\sum\limits_{n = 1}^m {\frac{1}{{2n - 1}}} } = }$$ $$\displaystyle{\sum\limits_{m = 1}^\infty {{{\left( { - 1} \right)}^m}\sum\limits_{n = 1}^m {\frac{1}{ {2n - 1}}} \int\limits_0^1 {{x^{2m}}dx} } = \int\limits_0^1 {\left( {\sum\limits_{m = 1}^\infty {{ {\left( { - 1} \right)}^m}\left( {\sum\limits_{n = 1}^m {\frac{1}{{2n - 1}}} } \right){x ^{2m}}} } \right)dx} }$$

We consider the function

$$\displaystyle{f\left( x \right) = \sum\limits_{m = 1}^\infty {{{\left( { - 1} \right)}^m}\left( {\sum\limits_{ n = 1}^m {\frac{1}{{2n - 1}}} } \right){x^{2m}}} = \sum\limits_{m = 1}^\infty {{a_m} \cdot {x^{2m}}} }$$ with $$\displaystyle{{a_m} = {\left( { - 1} \right)^m}\left( {\sum\limits_{n = 1}^m {\frac{1}{{2n - 1}}} } \right)}$$

Then $$\displaystyle{a_m} = {\left( { - 1} \right)^m}\left( {\sum\limits_{n = 1}^m {\frac{1}{{2n - 1}}} } \right) \Rightarrow {a_{m + 1}} = - {\left( { - 1} \right)^m}\left( {\sum\limits_{n = 1}^{m + 1} { \frac{1}{{2n - 1}}} } \right) = $$ $$\displaystyle{ - {\left( { - 1} \right)^m}\left( {\frac{1}{{2m + 1}} + \sum\limits_{n = 1}^m {\frac{ 1}{{2n - 1}}} } \right) = - {a_m} - \frac{{{{\left( { - 1} \right)}^m}}}{{2m + 1}}}$$

Answer 1

\begin{gathered} * {\text{This equality holds}}:\frac{{\arctan \left( x \right)}}{{1 + {x^2}}} = - \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}} \left( {{H_{2n}} - \frac{1}{2}{H_n}} \right){x^{2n - 1}}\left( {\left| x \right| \leqslant 1} \right) \hfill \\ \Leftrightarrow - \frac{{x\arctan \left( x \right)}}{{1 + {x^2}}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}} \left( {{H_{2n}} - \frac{1}{2}{H_n}} \right){x^{2n}} \hfill \\ {\text{Now, integrate both side w}}{\text{.r}}{\text{.t }}x{\text{ from }}0{\text{ to }}1: \hfill \\ \Rightarrow - \int\limits_0^1 {\frac{{x\arctan \left( x \right)}}{{1 + {x^2}}}dx} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}} \left( {{H_{2n}} - \frac{1}{2}{H_n}} \right)\int\limits_0^1 {{x^{2n}}} dx = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{2n + 1}}} \left( {{H_{2n}} - \frac{1}{2}{H_n}} \right) = S \hfill \\ * I = \int\limits_0^1 {\frac{{x\arctan \left( x \right)}}{{1 + {x^2}}}dx} ,{\text{ let}}:x = \tan \left( t \right) \Rightarrow dx = \left( {1 + {{\tan }^2}\left( t \right)} \right)dt \Rightarrow \frac{{dx}}{{1 + {x^2}}} = dt \hfill \\ \Rightarrow I = \int\limits_0^{\frac{\pi }{4}} {t\tan \left( t \right)dt} \xrightarrow{{{\text{IBP}}}}\left. { - t\ln \left( {\cos \left( t \right)} \right)} \right|_0^{\frac{\pi }{4}} + \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\cos \left( t \right)} \right)dt} = \frac{\pi }{8}\ln \left( 2 \right) + \frac{G}{2} - \frac{\pi }{4}\ln \left( 2 \right) = \frac{G}{2} - \frac{\pi }{8}\ln \left( 2 \right) \hfill \\ \because \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\cos \left( t \right)} \right)dt} = \frac{G}{2} - \frac{\pi }{4}\ln \left( 2 \right){\text{ is well}} - {\text{known, and can be found in this site}}{\text{.}} \hfill \\ \Rightarrow S = - I = \frac{\pi }{8}\ln \left( 2 \right) - \frac{G}{2} \hfill \\ \end{gathered}

Answer 2

I will continue your work

$$\displaystyle{ \Rightarrow {a_{m + 1}} \cdot {x^{2m}} = - {a_m} \cdot {x^{2m}} - \frac{{{{\left( { - 1} \right)}^m}}}{{2m + 1}} \cdot {x^{2m}} \Rightarrow \sum\limits_{m = 1}^\infty {{a_{m + 1}} \cdot {x^{2m}}} }$$ $$\displaystyle{ = - \sum\limits_{m = 1}^\infty {{a_m} \cdot {x^{2m}}} - \sum\limits_{m = 1}^\infty {\frac{{{{\left( { - 1} \right)}^m}}}{{2m + 1}} \cdot {x^{2m}}} \Rightarrow }$$

$$\displaystyle{ \Rightarrow \sum\limits_{m = 2}^\infty {{a_m} \cdot {x^{2m - 2}}} = - \sum\limits_{m = 1}^\infty {{a_m} \cdot {x^{2m}}} - \sum\limits_{m = 1}^\infty {\frac{{{{\left( { - 1} \right)}^m}}}{{2m + 1}} \cdot {x^{2m}}} \Rightarrow }$$ $$\displaystyle{\frac{1}{{{x^2}}}\sum\limits_{m = 2}^\infty {{a_m} \cdot {x^{2m}}} = - f\left( x \right) - \frac{1}{x}\sum\limits_{m = 1}^\infty {\frac{{{{\left( { - 1} \right)}^m}}}{{2m + 1}} \cdot {x^{2m + 1}}} \Rightarrow }$$

$$\displaystyle{ \Rightarrow \frac{1}{{{x^2}}}\left( { - {a_1} \cdot {x^2} + f\left( x \right)} \right) = - f\left( x \right) - \frac{1}{x}\left( { - x + \sum\limits_{m = 0}^\infty {\frac{{{{\left( { - 1} \right)}^m}}}{{2m + 1}} \cdot {x^{2m + 1}}} } \right)} \displaystyle{ = \frac{1}{{{x^2}}}\left( {{x^2} + f\left( x \right)} \right) = - f\left( x \right) - \frac{1}{x}\left( { - x + Arc\tan x} \right) \Rightarrow }$$

$$\displaystyle{ \Rightarrow f\left( x \right) = \sum\limits_{m = 1}^\infty {{{\left( { - 1} \right)}^m}\left( {\sum\limits_{n = 1}^m {\frac{1}{{2n - 1}}} } \right){x^{2m}}} = - \frac{{x \cdot Arc\tan x}}{{1 + {x^2}}}}$$

$$\displaystyle{S = - \int\limits_0^1 {\frac{{x \cdot Arc\tan x}}{{1 + {x^2}}}dx} = - \frac{1}{2}\int\limits_0^1 {Arc\tan x\left( {\log \left( {1 + {x^2}} \right)} \right)'dx} } \displaystyle{ = - \frac{{\pi \cdot \log 2}}{8} + \frac{1}{2} \cdot \int\limits_0^1 {\frac{{\log \left( {1 + {x^2}} \right)}}{{1 + {x^2}}}dx} \mathop { = = = }\limits^{x = \tan z} }$$

$$\displaystyle{ = - \frac{{\pi \cdot \log 2}}{8} + \frac{1}{2} \cdot \int\limits_0^{\pi /4} {\log \left( {1 + {{\tan }^2}u} \right)dx} = - \frac{{\pi \cdot \log 2}}{8} - \int\limits_0^{\pi /4} {\log \left( {\cos u} \right)dx} = }$$

$$\displaystyle{ = - \frac{{\pi \cdot \log 2}}{8} - \int\limits_0^{\pi /4} {\left( { - \log 2 + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\cos 2nx}}{n}} } \right)dx} } \displaystyle{ = \frac{{\pi \cdot \log 2}}{8} - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}\int\limits_0^{\pi /4} {\cos 2nx\;dx} } = }$$

$$\displaystyle{ = \frac{{\pi \cdot \log 2}}{8} - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{2{n^2}}} \cdot \sin \frac{{n\pi }}{2}} = \frac{{\pi \cdot \log 2}}{8}} \displaystyle{ - \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{{\left( {2n - 1} \right)}^2}}} \cdot {{\left( { - 1} \right)}^{n - 1}}} = \frac{{\pi \cdot \log 2}}{8} - \frac{G}{2}}$$ :)