n-1-n-n-n-

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

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

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

R A T I O T E S T

a_{n + 1 .} . \frac{1}{a_{n}} = \frac{(n + 1)!}{{(n + 1)}^{n + 1}} . \frac{n^{n}}{n!} = {(\frac{n}{n + 1})}^{n}

l i m {(\frac{1}{1 + \frac{1}{n}})}^{n} = {(1 + \frac{1}{n})}^{- n} = \frac{1}{e}

n \to \infty

S e c o n d m e t h o d

l i m n l o g (\frac{n}{n + 1}) = l o g y \Rightarrow n (\frac{n}{n + 1} - 1) = l o g y \Rightarrow - 1 = l o g y \Rightarrow y = \frac{1}{e}

n \to \infty

C o n v e r g e s

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

convergence ? let u_{n} = \frac{n!}{n^{n}} we have n! \sim n^{n} e^{- n} \sqrt{2 π n} (n \to \infty) \Rightarrow

u_{n} \sim e^{- n} \sqrt{2 π n} and Σ e^{- n} \sqrt{2 π n} is the same nature of \int_{0}^{\infty} e^{- t} \sqrt{2 π t} dt but

\int_{0}^{\infty} \sqrt{2 π} e^{- t} \sqrt{t} dt = \sqrt{2 π} \int_{0}^{\infty} t^{\frac{3}{2} - 1} e^{- t} dt = \sqrt{2 π} Γ (\frac{3}{2}) this integral convergez \Rightarrow

Σ u_{n} converges . .

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

C a n y o u f i n d t h e s u m s i r ?

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

perhaps by using some relation and formulas \dots