Question Number 134016 by mnjuly1970 last updated on 26/Feb/21
$$ \\ $$$$\:\:\:\:\:\:?{prove}\::\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {H}_{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right).. \\ $$
Answered by mnjuly1970 last updated on 27/Feb/21
![Σ_(n=1) ^∞ (((−1)^n H_(2n) )/(2n+1))=(1/2)Σ_(n=1) ^∞ (((i^n H_n )/(n+1)))+(1/2)Σ_(n=1) ^∞ ((((−1)^n i^n H_n )/(n+1))) =(1/2)Σ_(n=1) ^∞ {i^n H_n ∫_0 ^( 1) x^n dx}+(1/2)Σ_(n=) ^∞ {(−i)^n H_n ∫_0 ^( 1) x^n dx} =−(1/2)∫_0 ^( 1) ((ln(1−ix))/(1−ix))dx−(1/2)∫_0 ^( 1) ((ln(1+ix))/(1+ix))dx =(1/(4i))[ln^2 (1−xi)]_0 ^1 −(1/(4i))[ln^2 (1+xi)]_(0 ) ^1 =(1/(4i))(ln((√2) e^((−πi)/4) ))^2 −(1/(4i))(ln((√2) e^((πi)/4) ))^2 =(1/(4i))[(ln((√2) )−((iπ)/4))^2 −(ln(√(2))) +((iπ)/4))^2 ] =(1/(4i))(−4ln((√2) )(((iπ)/4)))=((−π)/8)ln(2) ∴ S=(π/8)ln(2) ...✓✓ ...m.n...](https://www.tinkutara.com/question/Q134063.png)
$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{i}^{{n}} {H}_{{n}} }{{n}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} {i}^{{n}} {H}_{{n}} }{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{{i}^{{n}} {H}_{{n}} \int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}} {dx}\right\}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=} {\overset{\infty} {\sum}}\left\{\left(−{i}\right)^{{n}} {H}_{{n}} \int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}} {dx}\right\} \\ $$$$\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{ix}\right)}{\mathrm{1}−{ix}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{ix}\right)}{\mathrm{1}+{ix}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\left[{ln}^{\mathrm{2}} \left(\mathrm{1}−{xi}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{4}{i}}\left[{ln}^{\mathrm{2}} \left(\mathrm{1}+{xi}\right)\right]_{\mathrm{0}\:} ^{\mathrm{1}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\left({ln}\left(\sqrt{\mathrm{2}}\:{e}^{\frac{−\pi{i}}{\mathrm{4}}} \right)\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}{i}}\left({ln}\left(\sqrt{\mathrm{2}}\:{e}^{\frac{\pi{i}}{\mathrm{4}}} \right)\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\left[\left({ln}\left(\sqrt{\mathrm{2}}\:\right)−\frac{{i}\pi}{\mathrm{4}}\right)^{\mathrm{2}} −\left({ln}\sqrt{\left.\mathrm{2}\right)}\:+\frac{{i}\pi}{\mathrm{4}}\right)^{\mathrm{2}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\left(−\mathrm{4}{ln}\left(\sqrt{\mathrm{2}}\:\right)\left(\frac{{i}\pi}{\mathrm{4}}\right)\right)=\frac{−\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\therefore\:{S}=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\:\:…\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:…{m}.{n}… \\ $$