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Question Number 214485 by depressiveshrek last updated on 09/Dec/24

lim_(n→∞) ((1/(2∙4))+(1/(5∙7))+...+(1/((3n−1)(3n+1))))

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}\centerdot\mathrm{7}}+...+\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{1}\right)}\right) \\ $$

Answered by mr W last updated on 10/Dec/24

ψ(1−(1/3))=−γ+Σ_(n=1) ^∞ ((1/n)−(1/(n−(1/3)))) ψ(1+(1/3))=−γ+Σ_(n=1) ^∞ ((1/n)−(1/(n+(1/3)))) ψ(1+(1/3))−ψ(1−(1/3))=Σ_(n=1) ^∞ ((1/(n−(1/3)))−(1/(n+(1/3)))) ψ((1/3))+3−ψ((1/3))−π cot (π/3)=Σ_(n=1) ^∞ ((1/(n−(1/3)))−(1/(n+(1/3)))) 3−(π/( (√3)))=Σ_(n=1) ^∞ ((1/(n−(1/3)))−(1/(n+(1/3)))) Σ_(n=1) ^∞ (1/((3n−1)(3n+1))) =(1/2)Σ_(n=1) ^∞ ((1/(3n−1))−(1/(3n+1))) =(1/6)Σ_(n=1) ^∞ ((1/(n−(1/3)))−(1/(n+(1/3)))) =(1/6)(3−(π/( (√3)))) =(1/2)−(π/(6(√3))) ✓

$$\psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=−\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)=−\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{3}−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\pi\:\mathrm{cot}\:\frac{\pi}{\mathrm{3}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$\mathrm{3}−\frac{\pi}{\:\sqrt{\mathrm{3}}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{3}−\frac{\pi}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}}\:\checkmark \\ $$

Commented by MathematicalUser2357 last updated on 12/Dec/24

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