Question Number 123075 by nico last updated on 22/Nov/20
Answered by Olaf last updated on 23/Nov/20
![((2n+2i+1)/(2(n+i)^2 )) = (1/(n+i))+(1/(2(n+i)^2 )) Σ_(i=1) ^n (1/(n+i)) = ψ(2n+1)−ψ(n+1) Σ_(i=1) ^n (1/((n+i)^2 )) = (1/2)[−ψ_1 (2n+1)+ψ_1 (n+1)] to be continued...](https://www.tinkutara.com/question/Q123105.png)
$$\frac{\mathrm{2}{n}+\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}\left({n}+{i}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{n}+{i}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+{i}\right)^{\mathrm{2}} } \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}+{i}}\:=\:\psi\left(\mathrm{2}{n}+\mathrm{1}\right)−\psi\left({n}+\mathrm{1}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left({n}+{i}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[−\psi_{\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)+\psi_{\mathrm{1}} \left({n}+\mathrm{1}\right)\right] \\ $$$${to}\:{be}\:{continued}… \\ $$