Question Number 140405 by mnjuly1970 last updated on 07/May/21
$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:{find}\:\:{the}\:\:{value}\:{of}\::: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\Theta\::=\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}.\left(\mathrm{4}{n}+\mathrm{1}\right).\left(\mathrm{4}{n}+\mathrm{2}\right).\left(\mathrm{4}{n}+\mathrm{3}\right)}=? \\ $$$$\:\:\:\:\: \\ $$
Answered by qaz last updated on 07/May/21
![f(n)=(1/(4n(4n+1)(4n+2)(4n+3)))=(A/(4n))+(B/(4n+1))+(C/(4n+2))+(D/(4n+3)) A=lim_(4n→0) 4nf(n)=(1/6) B=lim_(4n→−1) (4n+1)f(n)=−(1/2) C=lim_(4n→−2) (4n+2)f(n)=(1/2) D=lim_(4n→−3) (4n+3)f(n)=−(1/6) Θ=Σ_(n=1) ^∞ [(1/(6×4n))−(1/(2(4n+1)))+(1/(2(4n+2)))−(1/(6(4n+3)))] =Σ_(n=1) ^∞ [(1/(24n))−(1/(8(n+(1/4))))+(1/(8(n+(1/2))))−(1/(24(n+(3/4))))] =(1/(24))ψ((7/4))+(1/8)ψ((5/4))−(1/8)ψ((3/2)) =−(1/(24))γ+((11)/(36))−(π/(24))−(1/4)ln2](https://www.tinkutara.com/question/Q140422.png)
$${f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{4}{n}\left(\mathrm{4}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{2}\right)\left(\mathrm{4}{n}+\mathrm{3}\right)}=\frac{{A}}{\mathrm{4}{n}}+\frac{{B}}{\mathrm{4}{n}+\mathrm{1}}+\frac{{C}}{\mathrm{4}{n}+\mathrm{2}}+\frac{{D}}{\mathrm{4}{n}+\mathrm{3}} \\ $$$${A}=\underset{\mathrm{4}{n}\rightarrow\mathrm{0}} {\mathrm{lim}4}{nf}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{B}=\underset{\mathrm{4}{n}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\mathrm{4}{n}+\mathrm{1}\right){f}\left({n}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${C}=\underset{\mathrm{4}{n}\rightarrow−\mathrm{2}} {\mathrm{lim}}\left(\mathrm{4}{n}+\mathrm{2}\right){f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:{D}=\underset{\mathrm{4}{n}\rightarrow−\mathrm{3}} {\mathrm{lim}}\left(\mathrm{4}{n}+\mathrm{3}\right){f}\left({n}\right)=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Theta=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{6}×\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{4}{n}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{4}{n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{4}{n}+\mathrm{3}\right)}\right] \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{24}{n}}−\frac{\mathrm{1}}{\mathrm{8}\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)}+\frac{\mathrm{1}}{\mathrm{8}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\mathrm{24}\left({n}+\frac{\mathrm{3}}{\mathrm{4}}\right)}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}\psi\left(\frac{\mathrm{7}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{8}}\psi\left(\frac{\mathrm{5}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{8}}\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{24}}\gamma+\frac{\mathrm{11}}{\mathrm{36}}−\frac{\pi}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{2} \\ $$