Absolute Convergence of sigma from n=1 to infinity of n!/n^n

[grapz]

I'm stuck on this question.

Sigma from n=1 to infinity of n!/ n ^n

I tried using the ratio test, but it gives me a limit = 1 which is inconclusive.

Ok this is weird, the limit of the ratio test actually = 0, when i plug it onto the calculator, but this is what my work shows me

After doing the ratio test, u get lim n --> infinity n^n / ( n + 1 )^n = 1/ ( n+1/n)^n = 1/ ( 1 + 1/n) ^n = 1 / ( 1 + 0 )^n = 1

but if u do plug n = 100000000000 into calculator into the last step. 1/ ( 1 + 1/n ) ^n

the limit to that is 0. Why is this?

Edit, i forgot that ( 1 + 1/n ) ^n = e

sorry about that

 

[Dr. Flim-Flam]

Use ratio test: lim as n->inf.|[(n+1)!/(n+1)^(n+1)]*(n^n/n!)| = [n/(n+1)]^n .

Let y =lim as n->inf [n/(n+1)]^n. then ln(y) =lim as n->inf [n*ln(n/(n+1))] = -1

ln(y) = -1 implies y = 1/e <1, therefore absolute convergence.

 

[pka]

The following is a very usefull limit when using the root test: \(\displaystyle \left( {\frac{{\sqrt[n]{{n!}}}}{n}} \right) \to \frac{1}{e}\).
The root test is also known as Cauchy’s Test and works in some important cases where the ratio test fails.