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Question Number 158522 by amin96 last updated on 05/Nov/21
Σ_(n=2) ^∞ (((ζ(n)−n))/2^n )=?
$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(\zeta\left({n}\right)−{n}\right)}{\mathrm{2}^{{n}} }=? \\ $$
Answered by qaz last updated on 06/Nov/21
Σ_(n=2) ^∞ ((ζ(n)−n)/2^n )  =Σ_(n=2) ^∞ Σ_(k=1) ^∞ (1/((2k)^n ))−(3/2)  =Σ_(k=1) ^∞ ((1/((2k)^2 ))/(1−(1/(2k))))−(3/2)  =(1/2)Σ_(k=1) ^∞ ((1/(k−(1/2)))−(1/k))−(3/2)  =−(1/2)H_(−1/2) −(3/2)  =ln2−(3/2)
$$\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{n}\right)−\mathrm{n}}{\mathrm{2}^{\mathrm{n}} } \\ $$$$=\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2k}\right)^{\mathrm{n}} }−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\frac{\mathrm{1}}{\left(\mathrm{2k}\right)^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{k}−\frac{\mathrm{1}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{k}}\right)−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{−\mathrm{1}/\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\mathrm{ln2}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by amin96 last updated on 06/Nov/21
thanks sir. very good solution
$${thanks}\:{sir}.\:{very}\:{good}\:{solution} \\ $$

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