Question and Answers Forum

All Questions      Topic List

Set Theory Questions

Previous in All Question      Next in All Question      

Previous in Set Theory      Next in Set Theory      

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

$${thanks}\:{sir}.\:{very}\:{good}\:{solution} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com