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Question Number 141030 by iloveisrael last updated on 15/May/21

→⟨Σ_(n=1) ^∞ (1/(9n^2 +3n)) =? ⟩←

$$\:\:\:\:\:\:\:\:\:\:\rightarrow\langle\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{9}{n}^{\mathrm{2}} +\mathrm{3}{n}}\:=?\:\rangle\leftarrow \\ $$

Commented by EDWIN88 last updated on 15/May/21

= 1−ln (√3) −(π/(6(√3))) S=(1/3) Σ_(n=1) ^∞ ((1/n)−(3/(3n+1))) S= (1/3) Σ_(n=1) ^∞ ((1/n)−(1/(n+(1/3)))) S= (1/3) [ ψ((4/3))+γ ] S= (1/3) [ ψ((1/3))+3+γ ] S = (1/3) [ 3−(3/2)ln 3−(π/(2(√3))) ] S = 1−(1/2)ln 3 −(π/(6(√3)))

$$\:=\:\mathrm{1}−\mathrm{ln}\:\sqrt{\mathrm{3}}\:−\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$$\:\mathrm{S}=\frac{\mathrm{1}}{\mathrm{3}}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{3}}{\mathrm{3n}+\mathrm{1}}\right) \\ $$$$\:\mathrm{S}=\:\frac{\mathrm{1}}{\mathrm{3}}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$\:\mathrm{S}=\:\frac{\mathrm{1}}{\mathrm{3}}\:\left[\:\psi\left(\frac{\mathrm{4}}{\mathrm{3}}\right)+\gamma\:\right] \\ $$$$\:\mathrm{S}=\:\frac{\mathrm{1}}{\mathrm{3}}\:\left[\:\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{3}+\gamma\:\right] \\ $$$$\:\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\left[\:\mathrm{3}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:\right] \\ $$$$\:\mathrm{S}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}\:−\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}}\: \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 15/May/21

Σ_(n=1) ^∞ (1/(3n(3n+1)))=Σ_(n=1) ^∞ (1/(3n))−(1/(3n+1)) =(1/3)Σ_(n=1) ^∞ (1/n)−(1/(n+(1/3)))=(1/3)ψ((4/3))+(γ/3)

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{n}\left(\mathrm{3}{n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{n}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{3}}\psi\left(\frac{\mathrm{4}}{\mathrm{3}}\right)+\frac{\gamma}{\mathrm{3}} \\ $$

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