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

Σ_(n=1) ^∞ (n^n /(n!))

n=1nnn!

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!))=∞

forlargen:n!nnen2πnnnn!en2πnlimnen2πn=n=1nnn!=

Commented by Her_Majesty last updated on 03/Sep/20

...but Σ_(n=1) ^∞ ((n!)/n^n ) exists

...butn=1n!nnexists

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)

limn(an+1an)=limn((n+1)n+1(n+1)!×n!nn)=limn(n+1)n(n+1)n!(n+1)n!nn=limn(n+1n)n=limn(1+1n)n=e>1Σn=1nnn!=(divergent)

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