n-1-n-n-n-

Question Number 102822 by Dwaipayan Shikari last updated on 11/Jul/20

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{{n}^{{n}} } \\ $$

Commented by Dwaipayan Shikari last updated on 11/Jul/20

RATIO TEST a_(n+1.) .(1/a_n )=(((n+1)!)/((n+1)^(n+1) )).(n^n /(n!))=((n/(n+1)))^n lim((1/(1+(1/n))))^n =(1+(1/n))^(−n) =(1/e) n→∞ Second method lim nlog((n/(n+1)))=logy ⇒ n((n/(n+1))−1)=logy ⇒−1=logy⇒y=(1/e) n→∞ Converges

$${RATIO}\:{TEST} \\ $$$${a}_{{n}+\mathrm{1}.} .\frac{\mathrm{1}}{{a}_{{n}} }=\frac{\left({n}+\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }.\frac{{n}^{{n}} }{{n}!}=\left(\frac{{n}}{{n}+\mathrm{1}}\right)^{{n}} \\ $$$${lim}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\right)^{{n}} =\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{−{n}} =\frac{\mathrm{1}}{{e}} \\ $$$$ \\ $$$${n}\rightarrow\infty \\ $$$${Second}\:{method} \\ $$$${lim}\:{nlog}\left(\frac{{n}}{{n}+\mathrm{1}}\right)={logy}\:\:\Rightarrow\:{n}\left(\frac{{n}}{{n}+\mathrm{1}}−\mathrm{1}\right)={logy}\:\Rightarrow−\mathrm{1}={logy}\Rightarrow{y}=\frac{\mathrm{1}}{{e}} \\ $$$${n}\rightarrow\infty \\ $$$${Converges} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 11/Jul/20

convergence ? let u_n =((n!)/n^n ) we have n! ∼ n^n e^(−n) (√(2πn)) (n→∞) ⇒ u_n ∼e^(−n) (√(2πn)) and Σ e^(−n) (√(2πn))is the same nature of ∫_0 ^∞ e^(−t) (√(2πt))dt but ∫_0 ^∞ (√(2π)) e^(−t) (√t)dt =(√(2π))∫_0 ^∞ t^((3/2)−1) e^(−t) dt =(√(2π))Γ((3/2)) this integral convergez ⇒ Σ u_n converges..

$$\mathrm{convergence}\:?\:\:\mathrm{let}\:\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{n}!}{\mathrm{n}^{\mathrm{n}} }\:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{n}!\:\sim\:\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \:\sqrt{\mathrm{2}\pi\mathrm{n}}\:\:\left(\mathrm{n}\rightarrow\infty\right)\:\Rightarrow \\ $$$$\mathrm{u}_{\mathrm{n}} \sim\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\:\:\:\mathrm{and}\:\Sigma\:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{nature}\:\mathrm{of}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}} \sqrt{\mathrm{2}\pi\mathrm{t}}\mathrm{dt}\:\mathrm{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\sqrt{\mathrm{2}\pi}\:\mathrm{e}^{−\mathrm{t}} \:\sqrt{\mathrm{t}}\mathrm{dt}\:=\sqrt{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\sqrt{\mathrm{2}\pi}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:\mathrm{this}\:\mathrm{integral}\:\mathrm{convergez}\:\Rightarrow \\ $$$$\Sigma\:\mathrm{u}_{\mathrm{n}} \:\mathrm{converges}.. \\ $$

Commented by Dwaipayan Shikari last updated on 11/Jul/20

$${Can}\:{you}\:{find}\:{the}\:{sum}\:{sir}? \\ $$

Commented by mathmax by abdo last updated on 12/Jul/20

perhaps by using some relation and formulas...

$$\mathrm{perhaps}\:\mathrm{by}\:\mathrm{using}\:\mathrm{some}\:\mathrm{relation}\:\mathrm{and}\:\mathrm{formulas}… \\ $$

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