Radius of convergence of power series $\sum_0^{\infty} n!x^{n^2}$

Question

Radius of convergence of power series $\sum_0^{\infty} n!x^{n^2}$.

I applied ratio test, it went inconclusive and in root test i got $|x|^n n!^{1/n}$ but $n!^{1/n} $ diverges as n gets bigger.

So please help me to proceed

Answer 1

HINT:

Note that we have

$$ \limsup_{n\to\infty}\sqrt[n]{\left| n!x^{n^2}\right|}=\limsup_{n\to\infty}(n!)^{1/n}|x|^n$$

Now use Stirling' s Formula

$$n!\sim \sqrt{2\pi n}\left( \frac ne \right)^n$$

Answer 2

The power series is $\sum_0^\infty a_m x^m$ where $$ a_m = \begin{cases} n!, & m = n^2, n \in \mathbb N^*\\ 0, & \ \text{otherwise}, \end{cases} $$ so by Hadamard formula, $$ 1/R =\varlimsup |a_m|^{1/m} = \varlimsup (n!)^{1/n^2} =\varlimsup (\sqrt{2\pi} n^{n+1/2} \mathrm e^{-n})^{1/n^2} = \lim (2\pi)^{1/2n^2} n^{1/n} n^{1/2n^2}\mathrm e^{-1/n} = 1, $$ where we use the Stirling approximation. Hence the radius of convergence is $1$.

Answer 3

Your formula is wrong because the exponent of $x$ is $n^2,$ but you can use the Weak form of Stirling’s Formula \begin{align} \dfrac{n^n}{e^{n-1}}<n!<\dfrac{n^{n+1}}{e^{n-1}} . \end{align} This implies that \begin{align} \sum^{\infty}_{n=0} n!{x^{n^2}}\leq \sum^{\infty}_{n=0} \dfrac{n^{n+1}}{e^{n-1}} {x^{n^2}}. \end{align} So, fix $x\in \Bbb{R},$ then by D'Alembert's ratio test \begin{align} \lim\limits_{n\to \infty}\left|\dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} \dfrac{e^{n-1}} {n^{n+1} x^{n^2}}\right|&=\lim\limits_{n\to \infty}\left|\dfrac{n }{e} {x^{(n+1)^2-n^2}}\right|\\&=\lim\limits_{n\to \infty}\left|\dfrac{n }{e} {x^{2n+1}}\right|\\&=\dfrac{\left|x \right|}{e} \lim\limits_{n\to \infty}n\, {\left|x\right|}^{2n}. \end{align}

CASE I: $|x|<1$

Since $|x|<1$, then ${\left|x\right|}^{2n}\to 0$, and so, there exists $N\in \Bbb{N}$ such that \begin{align} {\left|x\right|}^{2n}<\dfrac{1}{n^2},\forall\;n\geq N .\end{align} Thus, \begin{align} \lim\limits_{n\to \infty}\left|\dfrac{n^{n+2}{x^{(n+1)^2}}}{e^{n}} \dfrac{e^{n-1}} {n^{n+1} x^{n^2}}\right|&=\dfrac{\left|x \right|}{e} \lim\limits_{n\to \infty}n\, {\left|x\right|}^{2n}\leq \dfrac{\left|x \right|}{e} \lim\limits_{n\to \infty}n\, \left(\dfrac{1}{n^2}\right)=0<1. \end{align} Hence, the series converges when $|x|<1.$

CASE II: $|x|\geq 1$
\begin{align} \lim\limits_{n\to \infty}n!{x^{n^2}}=\infty \end{align} which implies that the series cannot not converge. Consequently, the radius of convergence is $1$.