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Question-109141

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Question Number 109141 by mathdave last updated on 21/Aug/20
Answered by Dwaipayan Shikari last updated on 21/Aug/20
Σ_(n=1) ^∞ (1/(2^n n))=(1/2)+(1/2^2 ).(1/2)+(1/(2^3 .3))+....=−log(1−(1/2))=log(2)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} {n}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}}+….=−{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)={log}\left(\mathrm{2}\right) \\ $$
Commented by mathdave last updated on 21/Aug/20
the question u solved wasnt the question above
$${the}\:{question}\:{u}\:{solved}\:{wasnt}\:{the}\:{question}\:{above} \\ $$
Answered by Dwaipayan Shikari last updated on 21/Aug/20
S=Σ_(n=1) ^∞ (1/(2^n n^3 ))=(1/(2.1^3 ))+(1/(2^2 2^3 ))+(1/(2^3 .3^3 ))+.... (S/2)=(1/(2^2 .1^3 ))+(1/(2^3 .2^3 ))+.... ......subtracting (S/2)=(1/(2.1^3 ))+(1/2^2 )((1/2^3 )−1)+(1/2^3 )((1/2^3 )−(1/3^3 ))+.... (S/2)=(1/2)−((1/2^2 ).(7/(8.1))+(1/2^3 )+((19)/(27.8))+(1/2)+....) S=1−2S′ S′=(7/(8.1)).(1/2^2 )+((19)/(27.8)).(1/2^3 )+...continue
$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} {n}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} \mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{3}} }+…. \\ $$$$\frac{{S}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} .\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} .\mathrm{2}^{\mathrm{3}} }+…. \\ $$$$……{subtracting} \\ $$$$\frac{{S}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\right)+…. \\ $$$$\frac{{S}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}−\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }.\frac{\mathrm{7}}{\mathrm{8}.\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{19}}{\mathrm{27}.\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}+….\right) \\ $$$${S}=\mathrm{1}−\mathrm{2}{S}' \\ $$$${S}'=\frac{\mathrm{7}}{\mathrm{8}.\mathrm{1}}.\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{19}}{\mathrm{27}.\mathrm{8}}.\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+…{continue} \\ $$$$ \\ $$

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