Question

another fun series! (-2)^n * (n!/(n^n) ) as n goes from 1 to infinity...converging or diverging??..i cant use the ratio test rite?

Answer 1

(-1)^n * (2)^n * n!/n^N
consider the seq of positive term and apply the ratio test

Answer 2

you can use the ratio test here

Answer 3

\[\sum_{n=1}^{\infty} (-2)^n \frac{n!}{n^n}\]

applying the ratio test theorem , we'll get:
\[|\frac{an+1}{an}| = |\frac{(-2)^{n+1}(n+1)!}{(n+1)^{n+1}}|\]\[ \.\frac{n^n}{n!.(-2)^n}\]=
\[=\frac{-2^n.2^1.n!(n+1)}{(n+1)^n(n+1)}[\frac{n^n}{(-2)^n .n!}\]

Answer 4

\[=\frac{2n^n}{(n+1)^n}\]
after that find the limit and you'll compare your answers

If L < 1 = convergent
If L > 1 = divergent
if L = 1, no conclusion

^_^ L = limit, you can take it from here :)