Calculus
posted by andre
Calculate the interval of convergence.
Sigma from n=1 to infinity of (n!*x^n)/n^n.
Calculus  Steve, Friday, July 29, 2016 at 11:52pm
try using the ratio test. You should be able to show that the series converges for x < e
I tried and I got
xlim ninfinity of (n/n+1)^n
which I get x<1

Steve
Using the ratio test,
An+1/An =
(n+1)!/(n+1)^(n+1)
 x
n!/n^n
= (n+1)!/n! n^n/(n+1)^(n+1) x
= (n+1) (n/(n+1))^n * 1/(n+1) x
= (n/(n+1))^n x
Hmmm. what to do?
= ((n+1)/n)^n x
= (1 + 1/n)^n x
This should look familiar
(1 + 1/n)^n > 1/e
So, now we have the ratio test saying that
An+1/An = 1/e x
For the series to converge, the ratio must be less than 1, which means that
x < e
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