What is the interval of convergence of #sum_1^oo (x^n *n^n)/(n!)#?
1 Answer
Nov 5, 2015
Explanation:
Let
Then:
#a_(n+1) = (n+1)^(n+1)/((n+1)!) = (n+1)^n/(n!)=((n+1)/n)^n n^n/(n!)#
#=(1+1/n)^n a_n#
Now
So:
Hence if
If
#abs((x^(n+1)*(n+1)^(n+1))/((n+1)!) -: (x^n*n^n) / (n!)) > 1# for all#n >= N#
so the sum diverges.
I am not sure about the case