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Question Number 7771 by FilupSmith last updated on 14/Sep/16

Show that: Σ_(n=1) ^∞ (n/((n+2)!))=3−e

$$\mathrm{Show}\:\mathrm{that}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\left({n}+\mathrm{2}\right)!}=\mathrm{3}−{e} \\ $$

Commented by FilupSmith last updated on 14/Sep/16

Amazing!

$$\mathrm{Amazing}! \\ $$

Commented by sou1618 last updated on 14/Sep/16

S=Σ_(n=1) ^∞ (n/((n+2)!)) (1/((n+1)!))−(1/((n+2)!))=((n+1)/((n+2)!)) S=Σ_(n=1) ^∞ ((1/((n+1)!))−(2/((n+2)!))) g(x)=e^x g(x)=Σ_(i=0) ^∞ ((g^((i)) (0))/(i!))x^i g(1)=e=Σ_(i=0) ^∞ (1/(i!)) S=Σ_(n=0) ^∞ (1/((n+2)!))−Σ_(n=0) ^∞ (2/((n+3)!)) S={(Σ_(n=0) ^∞ (1/(n!)))−((1/(0!))+(1/(1!)))}−2{(Σ_(n=0) ^∞ (1/(n!)))−((1/(0!))+(1/(1!))+(1/(2!)))} 0!=1 S=(e−2)−2(e−(5/2)) S=3−e

$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\left({n}+\mathrm{2}\right)!} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!}=\frac{{n}+\mathrm{1}}{\left({n}+\mathrm{2}\right)!} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{2}}{\left({n}+\mathrm{2}\right)!}\right) \\ $$$$ \\ $$$${g}\left({x}\right)={e}^{{x}} \\ $$$${g}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{g}^{\left({i}\right)} \left(\mathrm{0}\right)}{{i}!}{x}^{{i}} \\ $$$${g}\left(\mathrm{1}\right)={e}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{i}!} \\ $$$$ \\ $$$${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)!}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{3}\right)!} \\ $$$${S}=\left\{\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\right)−\left(\frac{\mathrm{1}}{\mathrm{0}!}+\frac{\mathrm{1}}{\mathrm{1}!}\right)\right\}−\mathrm{2}\left\{\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\right)−\left(\frac{\mathrm{1}}{\mathrm{0}!}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}\right)\right\} \\ $$$$\mathrm{0}!=\mathrm{1} \\ $$$${S}=\left({e}−\mathrm{2}\right)−\mathrm{2}\left({e}−\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$${S}=\mathrm{3}−{e} \\ $$$$ \\ $$

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