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

Σ_(n=1) ^∞ ((n!)/n^n )

Commented byDwaipayan 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

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..

Commented byDwaipayan Shikari last updated on 11/Jul/20

Can you find the sum sir?

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

perhaps by using some relation and formulas...