Finding the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n!x^{n+9}}{(2n)!}$.

Question

I am having trouble understanding how to find the interval of convergence/radius. I know the interval of convergence is $(-\infty,\infty)$, and the radius is $\infty$ of $\sum_{n=0}^{\infty} \frac{n!x^{n+9}}{(2n)!}$. What I am having trouble with is actually finding the interval. I end up with $$\lim_{n\rightarrow \infty} \frac{1}{2(2n+1)} \times|x|$$ What I am having trouble with is conceptually understanding why it would converge for all value of $x$? What I did is that for it to converge $$ -1<\frac{1}{2(2n+1)} \times x<1$$ And then multiply all sides by $2(2n+1)$, but I'm having trouble understanding the finding the radius and interval process. All help is appreciated.

Answer 1

The $n$-th term of the power series is $a_n = \dfrac{n!x^{n+9}}{(2n)!}$.

Thus, $\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!x^{n+10}}{(2n+2)!} \cdot \dfrac{(2n)!}{n!x^{n+9}} = \dfrac{(n+1)x}{(2n+2)(2n+1)} = \dfrac{x}{2(2n+1)}$.

You did the above part just fine. Now, you need to correctly use the ratio test.

You should first evaluate the limit $\displaystyle\lim_{n \to \infty}\left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\dfrac{|x|}{2(2n+1)}$ (possibly in terms of $x$).

Then, determine for what values of $x$ is the limit between $-1$ and $1$ exclusive.