Question

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

Answer 1

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`