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

Jan 20, 2017

The radius of convergence is $R = \sqrt{e}$

#### Explanation:

We can apply the ratio test, stating that a series:

${\sum}_{n = 1}^{\infty} {a}_{n}$

is absolutely convergent if:

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid < 1$

Calculate the expression of the ratio for this series:

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

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

Now we have:

${\lim}_{n \to \infty} {\left(\frac{n}{n + 1}\right)}^{n} = {\lim}_{n \to \infty} \frac{1}{{\left(\frac{n + 1}{n}\right)}^{n}} = {\lim}_{n \to \infty} \frac{1}{{\left(1 + \frac{1}{n}\right)}^{n}} = \frac{1}{e}$

so that:

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {x}^{2} / e$

The series is then absolutely convergent if:

${x}^{2} / e < 1 \implies \left\mid x \right\mid < \sqrt{e}$

which means the radius of convergence is $R = \sqrt{e}$