How do you find the radius of convergence #Sigma (n^(2n))/((2n)!) x^n# from #n=[1,oo)#?

1 Answer
Andrea S.
Jan 11, 2017

The radius of convergence of the series:

#sum_(n=1)^oo n^(2n)/((2n)!)x^n#

is #R=4/e^2#

Explanation:

We have the series:

#sum_(n=1)^oo a_n = sum_(n=1)^oo n^(2n)/((2n)!)x^n#

We can apply the ratio test, to verify for which values of #x#, if any, we have:

#lim_(n->oo) abs(a_(n+1)/a_n) < 1#

#a_(n+1)/a_n = ((n+1)^(2(n+1))/((2(n+1))!)x^(n+1)) /(n^(2n)/((2n)!)x^n)#

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:

#a_(n+1)/a_n = (n+1)^2/(2(n+1)(2n+1)) ((n+1)/n)^(2n)x = (n+1)/(2(2n+1)) (1+1/n)^(2n)x #

Let us analyze the factors one by one:

#lim_(n->oo) (n+1)/(2(2n+1)) = 1/4#

#lim_(n->oo) (1+1/n)^(2n) = lim_(n->oo) ((1+1/n) ^n )^2 = e^2#

Finally we have:

#lim_(n->oo) abs (a_(n+1)/a_n) = e^2/4 abs(x) < 1 => abs(x) < 4/e^2#

so the radius of convergence is: #R=4/e^2#

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