Question Number 7771 by FilupSmith last updated on 14/Sep/16
$$\mathrm{Show}\:\mathrm{that}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\left({n}+\mathrm{2}\right)!}=\mathrm{3}−{e} \\ $$
Commented by FilupSmith last updated on 14/Sep/16
$$\mathrm{Amazing}! \\ $$
Commented by sou1618 last updated on 14/Sep/16
$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\left({n}+\mathrm{2}\right)!} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!}=\frac{{n}+\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{2}}{\left({n}+\mathrm{2}\right)!}\right) \\ $$$$ \\ $$$${g}\left({x}\right)={e}^{{x}} \\ $$$${g}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{g}^{\left({i}\right)} \left(\mathrm{0}\right)}{{i}!}{x}^{{i}} \\ $$$${g}\left(\mathrm{1}\right)={e}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{i}!} \\ $$$$ \\ $$$${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{3}\right)!} \\ $$$${S}=\left\{\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\right)−\left(\frac{\mathrm{1}}{\mathrm{0}!}+\frac{\mathrm{1}}{\mathrm{1}!}\right)\right\}−\mathrm{2}\left\{\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\right)−\left(\frac{\mathrm{1}}{\mathrm{0}!}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}\right)\right\} \\ $$$$\mathrm{0}!=\mathrm{1} \\ $$$${S}=\left({e}−\mathrm{2}\right)−\mathrm{2}\left({e}−\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$${S}=\mathrm{3}−{e} \\ $$$$ \\ $$