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Question Number 111466 by Dwaipayan Shikari last updated on 03/Sep/20

Σ_(n=1) ^∞ (n^n /(n!))

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

Commented by Her_Majesty last updated on 03/Sep/20

for large n: n!≈n^n e^(−n) (√(2πn))  ⇒ (n^n /(n!))≈(e^n /( (√(2πn))))  lim_(n→∞) (e^n /( (√(2πn))))=∞ ⇒ Σ_(n=1) ^∞ (n^n /(n!))=∞

$${for}\:{large}\:{n}:\:{n}!\approx{n}^{{n}} {e}^{−{n}} \sqrt{\mathrm{2}\pi{n}} \\ $$$$\Rightarrow\:\frac{{n}^{{n}} }{{n}!}\approx\frac{{e}^{{n}} }{\:\sqrt{\mathrm{2}\pi{n}}} \\ $$$${lim}_{{n}\rightarrow\infty} \frac{{e}^{{n}} }{\:\sqrt{\mathrm{2}\pi{n}}}=\infty\:\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{{n}} }{{n}!}=\infty \\ $$

Commented by Her_Majesty last updated on 03/Sep/20

...but Σ_(n=1) ^∞ ((n!)/n^n ) exists

$$...{but}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{{n}^{{n}} }\:{exists} \\ $$

Answered by malwaan last updated on 03/Sep/20

lim_(n→∞) ((a_(n+1) /a_n ))  = lim_(n→∞)  ((((n+1)^(n+1) )/((n+1)!)) × ((n!)/n^n ))  = lim_(n→∞)  (((n+1)^n (n+1)n!)/((n+1)n! n^n ))  = lim_(n→∞)  (((n+1)/n))^n   = lim_(n→∞)  (1+(1/n))^n  = e >1  ⇒Σ_(n=1) ^(∞)  (n^n /(n!)) = ∞ (divergent)

$$\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\right) \\ $$$$=\:\underset{{n}\rightarrow\infty} {{lim}}\:\left(\frac{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!}\:×\:\frac{{n}!}{{n}^{{n}} }\right) \\ $$$$=\:\underset{{n}\rightarrow\infty} {{lim}}\:\frac{\left({n}+\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right){n}!}{\left({n}+\mathrm{1}\right){n}!\:{n}^{{n}} } \\ $$$$=\:\underset{{n}\rightarrow\infty} {{lim}}\:\left(\frac{{n}+\mathrm{1}}{{n}}\right)^{{n}} \\ $$$$=\:\underset{{n}\rightarrow\infty} {{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \:=\:{e}\:>\mathrm{1} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\Sigma}}\:\frac{{n}^{{n}} }{{n}!}\:=\:\infty\:\left({divergent}\right) \\ $$

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