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

Feb 1, 2017

The series:

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

has radius of convergence $R = 1$

#### Explanation:

We can apply the ratio test:

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

now consider:

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

As ${e}^{x}$ is continuous:

${\lim}_{n \to \infty} {\left(n + 1\right)}^{\frac{1}{n}} = {e}^{{\lim}_{n \to \infty} \left(\ln \frac{n + 1}{n}\right)} = {e}^{0} = 1$

So we have:

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

which means that the series is absolutely convergent for $\left\mid x \right\mid < 1$ and divergent for $\left\mid x \right\mid > 1$, that is the radius of convergence is $R = 1$