Question Number 128740 by bemath last updated on 10/Jan/21
$$\:\:\underset{\mathrm{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{3}\right)\mathrm{n}!}\:=?\: \\ $$
Answered by liberty last updated on 10/Jan/21
$$\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{n}+\mathrm{2}}{\left(\mathrm{n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{1}\right)\mathrm{n}!}\:=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{n}+\mathrm{2}}{\left(\mathrm{n}+\mathrm{3}\right)!} \\ $$$$\:=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{n}+\mathrm{3}\right)−\mathrm{1}}{\left(\mathrm{n}+\mathrm{3}\right)!}\:=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)!}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{3}\right)!}\right)\: \\ $$$$\:\mathrm{Telescoping}\:\mathrm{series}\:=\:\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\mathrm{N}} {\sum}}\:\left(\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)!}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{3}\right)!}\right) \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$