Question Number 198372 by cortano12 last updated on 18/Oct/23
$$\:\:\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{4}×\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{7}×\mathrm{11}}\:+\frac{\mathrm{1}}{\mathrm{10}×\mathrm{14}}\:+\:\ldots=? \\ $$
Commented by Frix last updated on 19/Oct/23
$$\frac{\mathrm{1}}{\mathrm{8}}+\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{36}} \\ $$
Answered by mr W last updated on 19/Oct/23
![a_n =(1/((3n−2)(3n+2))) =(1/4)((1/(3n−2))−(1/(3n+2))) =(1/(12))((1/(n−(2/3)))−(1/(n+(2/3)))) Σ_(n=1) ^∞ a_n =(1/(12))(Σ_(n=1) ^∞ (1/(n−(2/3)))−Σ_(n=1) ^∞ (1/(n+(2/3)))) =(1/(12))[ψ(1+(2/3))−ψ(1−(2/3))] =(1/(12))[ψ((2/3))+(3/2)−ψ((1/3))] =(1/(12))[ψ((1/3))+π cot (π/3)+(3/2)−ψ((1/3))] =(1/(12))(π cot (π/3)+(3/2)) =(1/(12))((((√3)π)/3)+(3/2)) =(((√3)π)/(36))+(1/8)](https://www.tinkutara.com/question/Q198389.png)
$${a}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{12}}\left(\frac{\mathrm{1}}{{n}−\frac{\mathrm{2}}{\mathrm{3}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{2}}{\mathrm{3}}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} =\frac{\mathrm{1}}{\mathrm{12}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\frac{\mathrm{2}}{\mathrm{3}}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{\mathrm{2}}{\mathrm{3}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left[\psi\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left[\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\frac{\mathrm{3}}{\mathrm{2}}−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left[\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\pi\:\mathrm{cot}\:\frac{\pi}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}}−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\pi\:\mathrm{cot}\:\frac{\pi}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{36}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Commented by cortano12 last updated on 19/Oct/23
$$\mathrm{great} \\ $$