Question Number 134984 by Dwaipayan Shikari last updated on 09/Mar/21
$$\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{3}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{4}} }+... \\ $$$$ \\ $$Find in a closed form
Answered by Ñï= last updated on 09/Mar/21
![Σ_(n=1) ^∞ (H_n ^2 /2^n )=(1/(1−(1/2)))[ln^2 (1−(1/2))+Li_2 ((1/2))]=2ln^2 2+2Li_2 ((1/2))=ln^2 2+(π^2 /6) H_n :harmonic number](Q135007.png)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} ^{\mathrm{2}} }{\mathrm{2}^{{n}} }=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\left[{ln}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right]=\mathrm{2}{ln}^{\mathrm{2}} \mathrm{2}+\mathrm{2}{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)={ln}^{\mathrm{2}} \mathrm{2}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${H}_{{n}} :{harmonic}\:{number} \\ $$
Commented by Dwaipayan Shikari last updated on 09/Mar/21
$${Great}\:{sir}! \\ $$
Answered by mnjuly1970 last updated on 09/Mar/21
Commented by Dwaipayan Shikari last updated on 09/Mar/21
$${Great}\:{sir}!\:{Thanking}\:{you}! \\ $$
Commented by mnjuly1970 last updated on 09/Mar/21
$$\:{grateful}... \\ $$