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Prove-that-i-n-0-n-2-n-2e-ii-n-0-n-3-n-5e-iii-n-0-n-4-n-15e-

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Question Number 24540 by Tinkutara last updated on 20/Nov/17
Prove that (i) Σ_(n=0) ^∞ (n^2 /(n!))=2e. (ii) Σ_(n=0) ^∞ (n^3 /(n!))=5e. (iii) Σ_(n=0) ^∞ (n^4 /(n!))=15e.
$${Prove}\:{that} \\ $$$$\left({i}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{{n}!}=\mathrm{2}{e}. \\ $$$$\left({ii}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{{n}!}=\mathrm{5}{e}. \\ $$$$\left({iii}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{4}} }{{n}!}=\mathrm{15}{e}. \\ $$
Commented by prakash jain last updated on 20/Nov/17
Write a polynomial in n P(n) of degree k as P(n)=a_0 +a_1 n+a_2 n(n−1)+..+a_k n(n−1)...(n−k+1) solve for a_0 ,..a_k and continue as in case of n^2 in answer
$$\mathrm{Write}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{in}\:{n}\:{P}\left({n}\right)\:{of} \\ $$$${degree}\:{k}\:{as} \\ $$$${P}\left({n}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {n}+{a}_{\mathrm{2}} {n}\left({n}−\mathrm{1}\right)+..+{a}_{{k}} {n}\left({n}−\mathrm{1}\right)…\left({n}−{k}+\mathrm{1}\right) \\ $$$${solve}\:{for}\:{a}_{\mathrm{0}} ,..{a}_{{k}} \\ $$$${and}\:{continue}\:{as}\:{in}\:{case}\:{of}\:{n}^{\mathrm{2}} \:{in}\:{answer} \\ $$
Answered by prakash jain last updated on 20/Nov/17
n^2 =n+n(n−1) Σ_(n=0) ^∞ ((n+n(n−1))/(n!)) =Σ_(n=1) ^∞ (1/((n−1)!))+Σ_(n=2) ^∞ (1/((n−2)!)) =e+e=2e
$${n}^{\mathrm{2}} ={n}+{n}\left({n}−\mathrm{1}\right) \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}+{n}\left({n}−\mathrm{1}\right)}{{n}!} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{2}\right)!} \\ $$$$={e}+{e}=\mathrm{2}{e} \\ $$
Commented by Tinkutara last updated on 21/Nov/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Answered by ajfour last updated on 20/Nov/17
(i) xe^x =x+(x^2 /(1!))+(x^3 /(2!))+(x^4 /(3!))+.... (d/dx)(xe^x ) = 1 +((2x)/(1!)) + ((3x^2 )/(2!))+ ((4x^3 )/(3!))+... ⇒ e^x +xe^x =(1^2 /(1!))+((2^2 x)/(2!))+((3^2 x^2 )/(3!))+((4^2 x^3 )/(4!))+.. With x=1 we get 2e = Σ_(n=0) ^∞ (n^2 /(n!)) .
$$\left(\boldsymbol{{i}}\right)\:\:\:\:\:{xe}^{{x}} ={x}+\frac{{x}^{\mathrm{2}} }{\mathrm{1}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{3}!}+…. \\ $$$$\:\:\:\:\:\frac{{d}}{{dx}}\left({xe}^{{x}} \right)\:=\:\mathrm{1}\:+\frac{\mathrm{2}{x}}{\mathrm{1}!}\:\:+\:\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}!}+\:\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\Rightarrow\:{e}^{{x}} +{xe}^{{x}} \:=\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}!}+\frac{\mathrm{2}^{\mathrm{2}} {x}}{\mathrm{2}!}+\frac{\mathrm{3}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{\mathrm{4}^{\mathrm{2}} {x}^{\mathrm{3}} }{\mathrm{4}!}+.. \\ $$$${With}\:\boldsymbol{{x}}=\mathrm{1}\:\:{we}\:{get} \\ $$$$\:\:\:\:\mathrm{2}\boldsymbol{{e}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{2}} }{{n}!}\:. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 21/Nov/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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