n-1-1-n-2-2n-1-3-

Question Number 158531 by amin96 last updated on 05/Nov/21

$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} \left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)^{\mathrm{3}} }=? \\ $$

Answered by qaz last updated on 06/Nov/21

Σ_(n=1) ^∞ (1/(n^2 (2n+1)^3 )) =Σ_(n=1) ^∞ ((1/(8n^2 ))−(3/(4n))+(1/(2(2n+1)^3 ))+(1/((2n+1)^2 ))+(3/(2(2n+1)))) =(1/8)ζ(2)+(7/(16))ζ(3)+(3/4)ζ(2)+(3/4)Σ_(n=1) ^∞ ((1/(n+(1/2)))−(1/n)) =(7/8)ζ(2)+(7/(16))ζ(3)−(3/4)H_(1/2) =(7/8)ζ(2)+(7/(16))ζ(3)−(3/2)+(3/2)ln2

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} \left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{8n}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4n}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{2n}+\mathrm{1}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{2}\right)+\frac{\mathrm{7}}{\mathrm{16}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)+\frac{\mathrm{3}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{n}}\right) \\ $$$$=\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{2}\right)+\frac{\mathrm{7}}{\mathrm{16}}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{H}_{\mathrm{1}/\mathrm{2}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{2}\right)+\frac{\mathrm{7}}{\mathrm{16}}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln2} \\ $$

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