How do you find the radius of convergence #Sigma (n^(2n))/((2n)!) x^n# from #n=[1,oo)#?
1 Answer
The radius of convergence of the series:
is
Explanation:
We have the series:
We can apply the ratio test, to verify for which values of
Now we note that:
(i)
#x^(n+1)/x^n = x# (ii)
# ((2n)!)/((2(n+1))!) = ((2n)!)/((2n+2)(2n+1)((2n)!)) = 1/((2n+2)(2n+1)) = 1/(2(n+1)(2n+1))# (iii)
# (n+1)^(2(n+1))/n^(2n) = (n+1)^2(n+1)^(2n)/n^(2n) = (n+1)^2 ((n+1)/n)^(2n)#
So that we can express the ratio as:
Let us analyze the factors one by one:
Finally we have:
so the radius of convergence is: