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
|x|lim n-infinity of |(n/n+1)^n|
which I get |x|<1
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
To find the interval of convergence of the series sigma from n=1 to infinity of (n!*x^n)/n^n, you can use the ratio test.
The ratio test states that for a series sigma from n=1 to infinity of a_n, if the limit of |a_(n+1)/a_n| as n approaches infinity is L, then the series converges if L < 1 and diverges if L > 1.
Let's apply the ratio test to your series.
First, let's find the term a_(n+1)/a_n:
a_(n+1)/a_n = ((n+1)! * x^(n+1))/(n+1)^(n+1) * (n^n)/(n!)
= x * (n+1)/((n+1)^(n+1) * n^n/n!)
Now, let's take the limit of this expression as n approaches infinity:
lim n->infinity of (a_(n+1)/a_n) = lim n->infinity of (x * (n+1)/((n+1)^(n+1) * n^n/n!))
To simplify this limit, we can use the fact that n^n/n! approaches e as n approaches infinity. So, we replace n^n/n! with e:
lim n->infinity of (a_(n+1)/a_n) = lim n->infinity of (x * (n+1)/((n+1)^(n+1) * e))
Now, let's simplify further:
lim n->infinity of (x * (n+1)/((n+1)^(n+1) * e)) = lim n->infinity of (x/(n+1) * ((n+1)/(n+1))^(n+1) * e)
The expression ((n+1)/(n+1))^(n+1) approaches e as n approaches infinity. So, we can replace ((n+1)/(n+1))^(n+1) with e:
lim n->infinity of (x/(n+1) * e) = x * lim n->infinity of (e/(n+1))
Now, we can evaluate this limit:
lim n->infinity of (e/(n+1)) = 0
So, we have:
lim n->infinity of (a_(n+1)/a_n) = x * 0 = 0
Since the limit is 0, which is less than 1, we can conclude that the series converges for any value of x.
Therefore, the interval of convergence for the series sigma from n=1 to infinity of (n!*x^n)/n^n is (-infinity, +infinity).