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Question Number 78941 by M±th+et£s last updated on 21/Jan/20

Q.find the sum S=(2^3 /(2!))+(3^3 /(3!))+(4^3 /(4!))+.... then find Σ_(n=1) ^∞ (n^4 /(n!))

$${Q}.{find}\:{the}\:{sum} \\ $$$${S}=\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{4}!}+.... \\ $$$${then}\:{find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{4}} }{{n}!} \\ $$

Answered by mind is power last updated on 21/Jan/20

=Σ_(n≥1) (n^3 /((n−1)!))=Σ_(k≥0) (((k+1)^3 )/(k!))=Σ_(k≥0) (((k^3 +3k^2 +3k+1))/(k!)) =Σ_(k≥1) (k^2 /((k−1)!))+3Σ_(k≥1) (k/((k−1)!))+3Σ_(k≥1) (1/((k−1)!))+Σ_(k≥1) (1/(k!)) =Σ_(k≥0) (((k+1)^2 )/(k!))+3Σ_(k≥0) ((k+1)/(k!))+3Σ_(k≥0) (1/(k!))+Σ_(k≥0) (1/(k!))−1 =Σ_(k≥0) ((k^2 +2k+1)/(k!))+3Σ_(k≥0) (k/(k!))+Σ_(k≥0) (7/(k!))−1 =Σ_(k≥1) (k/((k−1)!))+2Σ_(k≥1) (1/((k−1)!))+Σ_(k≥0) (1/(k!))+3Σ_(k≥1) (1/((k−1)!))+7Σ_(k≥0) (1/(k!))−1 =Σ_(k≥0) ((k+1)/(k!))+2Σ_(k≥0) (1/(k!))+Σ_(k≥0) (1/(k!))+10Σ_(k≥0) (1/(k!))−1 =Σ_(k≥1) (1/((k−1)!))+Σ_(k≥0) (1/(k!))+2Σ_(k≥0) (1/(k!))+Σ_(k≥0) (1/(k!))+10Σ_(k≥0) (1/(k!))−1 =15Σ_(k≥0) (1/(k!))−1=15e−1

$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{n}^{\mathrm{3}} }{\left({n}−\mathrm{1}\right)!}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }{{k}!}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left({k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}\right)}{{k}!} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{k}^{\mathrm{2}} }{\left({k}−\mathrm{1}\right)!}+\mathrm{3}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{k}}{\left({k}−\mathrm{1}\right)!}+\mathrm{3}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}!} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{{k}!}+\mathrm{3}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}+\mathrm{1}}{{k}!}+\mathrm{3}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}−\mathrm{1} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}}{{k}!}+\mathrm{3}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}}{{k}!}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{7}}{{k}!}−\mathrm{1} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{k}}{\left({k}−\mathrm{1}\right)!}+\mathrm{2}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}+\mathrm{3}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}+\mathrm{7}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}−\mathrm{1} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}+\mathrm{1}}{{k}!}+\mathrm{2}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}+\mathrm{10}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}−\mathrm{1} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}+\mathrm{2}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}+\mathrm{10}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}−\mathrm{1} \\ $$$$=\mathrm{15}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}−\mathrm{1}=\mathrm{15}{e}−\mathrm{1} \\ $$

Answered by Smail last updated on 22/Jan/20

Σ_(n=1) ^∞ (n^4 /(n!))=Σ_(n=1) ^∞ (n^3 /((n−1)(n−2)(n−3)(n−4)!)) (n^3 /((n−1)!))=(1/((n−1)!))+(7/((n−2)!))+(6/((n−3)!))+(1/((n−4)!)) S=Σ_(n=4) ^∞ ((1/((n−1)!))+(7/((n−2)!))+(6/((n−3)!))+(1/((n−4)!)))+(1+((16)/2)+((27)/2)) S=((45)/2)+Σ_(n=4) ^∞ (1/((n−1)!))+Σ_(n=4) ^∞ (7/((n−2)!))+Σ_(n=4) ^∞ (6/((n−3)!))+Σ_(n=4) ^∞ (1/((n−4)!)) S=((45)/2)+Σ_(n=3) ^∞ (1/(n!))+Σ_(n=2) ^∞ (7/(n!))+Σ_(n=1) ^∞ (6/(n!))+Σ_(n=0) ^∞ (1/(n!)) =((45)/2)+(e−1−1−(1/2))+7(e−1−1)+6(e−1)+e =((45)/2)+15e−(5/2)−14−6 Σ_(n=1) ^∞ (n^4 /(n!))=15e

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{4}} }{{n}!}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)!} \\ $$$$\frac{{n}^{\mathrm{3}} }{\left({n}−\mathrm{1}\right)!}=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}+\frac{\mathrm{7}}{\left({n}−\mathrm{2}\right)!}+\frac{\mathrm{6}}{\left({n}−\mathrm{3}\right)!}+\frac{\mathrm{1}}{\left({n}−\mathrm{4}\right)!} \\ $$$${S}=\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}+\frac{\mathrm{7}}{\left({n}−\mathrm{2}\right)!}+\frac{\mathrm{6}}{\left({n}−\mathrm{3}\right)!}+\frac{\mathrm{1}}{\left({n}−\mathrm{4}\right)!}\right)+\left(\mathrm{1}+\frac{\mathrm{16}}{\mathrm{2}}+\frac{\mathrm{27}}{\mathrm{2}}\right) \\ $$$${S}=\frac{\mathrm{45}}{\mathrm{2}}+\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}+\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{7}}{\left({n}−\mathrm{2}\right)!}+\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{6}}{\left({n}−\mathrm{3}\right)!}+\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{4}\right)!} \\ $$$${S}=\frac{\mathrm{45}}{\mathrm{2}}+\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{7}}{{n}!}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{6}}{{n}!}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!} \\ $$$$=\frac{\mathrm{45}}{\mathrm{2}}+\left({e}−\mathrm{1}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{7}\left({e}−\mathrm{1}−\mathrm{1}\right)+\mathrm{6}\left({e}−\mathrm{1}\right)+{e} \\ $$$$=\frac{\mathrm{45}}{\mathrm{2}}+\mathrm{15}{e}−\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{14}−\mathrm{6} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{4}} }{{n}!}=\mathrm{15}{e} \\ $$

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