Question Number 211558 by mnjuly1970 last updated on 12/Sep/24
$$ \\ $$$$ \\ $$$$\:\:\:\:\:−−−−−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\Omega}=\:\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{n}}+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{n}}+\mathrm{1}}\:\right)=\:\boldsymbol{{a}\pi} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\boldsymbol{{a}}^{\mathrm{2}} =\:? \\ $$$$\:\:\:\:\:\:\:−−−−−−−−−−−− \\ $$
Answered by mr W last updated on 29/Sep/24
![(1/3)ψ(1+(2/3))=−(γ/3)+Σ_(n=1) ^∞ ((1/(3n))−(1/(3n+2))) (1/3)ψ(1+(1/3))=−(γ/3)+Σ_(n=1) ^∞ ((1/(3n))−(1/(3n+1))) (1/3)[ψ(1+(1/3))−ψ(1+(2/3))]=Σ_(n=1) ^∞ ((1/(3n+2))−(1/(3n+1))) (1/3)[ψ(1+(1/3))−ψ(1+(2/3))]=Σ_(n=0) ^∞ ((1/(3n+2))−(1/(3n+1)))+(1/2) (1/3)[ψ(1+(1/3))−ψ(1+(2/3))]−(1/2)=Σ_(n=0) ^∞ ((1/(3n+2))−(1/(3n+1))) (1/3)[ψ((1/3))+3−ψ((2/3))−(3/2)]−(1/2)=Σ_(n=0) ^∞ ((1/(3n+2))−(1/(3n+1))) (1/3)[ψ((1/3))−ψ(1−(1/3))]=Σ_(n=0) ^∞ ((1/(3n+2))−(1/(3n+1))) (1/3)[ψ((1/3))−ψ((1/3))−π cot (π/3)]=Σ_(n=0) ^∞ ((1/(3n+2))−(1/(3n+1))) −(π/(3(√3)))=Σ_(n=0) ^∞ ((1/(3n+2))−(1/(3n+1)))=aπ ⇒a=−(1/(3(√3))) ⇒a^2 =(1/(27)) ✓](https://www.tinkutara.com/question/Q212091.png)
$$\frac{\mathrm{1}}{\mathrm{3}}\psi\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)=−\frac{\gamma}{\mathrm{3}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)=−\frac{\gamma}{\mathrm{3}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left[\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\psi\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)\right]=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left[\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\psi\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)\right]=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left[\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\psi\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)\right]−\frac{\mathrm{1}}{\mathrm{2}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left[\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{3}−\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\frac{\mathrm{3}}{\mathrm{2}}\right]−\frac{\mathrm{1}}{\mathrm{2}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left[\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\right]=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left[\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\pi\:\mathrm{cot}\:\frac{\pi}{\mathrm{3}}\right]=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right) \\ $$$$−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}\right)={a}\pi \\ $$$$\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{27}}\:\checkmark \\ $$