My question is: how do I calculate the radius convergence of a power series when the series is not written like
$$\sum a_{n}x^{n}?$$
I have this series:
$$\sum\frac{x^{2n+1}}{(-3)^{n}}$$
Can I use the criterions as I was working with $x^{n}$, not $x^{2n+1}$?
I tried this:
$$k=2n+1\Rightarrow n=\frac{k-1}{2}$$
And I got
$$R=\lim_{k\to\infty}\left|\frac{a_{k}}{a_{k+1}}\right|=\frac{1}{\sqrt{3}}$$
I know the answer is: the series converges for all $x$ that $|x|<\sqrt{3}$. How do I get it?
Thanks 🙂
Best Answer
You can still use the ratio test; suppose $x\ne0$; then the ratio to compute is $$ \frac{|x|^{2n+3}}{3^{n+1}}\bigg/\frac{|x|^{2n+1}}{3^{n}}= \frac{|x|^{2n+3}}{3^{n+1}}\frac{3^{n}}{|x|^{2n+1}}=\frac{|x|^2}{3} $$ This is constant, so the series is actually a geometric series, easy to analyze; but, also in general, you know that, as long as the limit of the ratios is less than $1$, the series is convergent.
This is not always applicable, but if the limit exists, you can conclude.
For a different example, consider $$ \sum_{n\ge0}\frac{x^{2n+1}}{(2n+1)!} $$ where the ratio to compute is $$ \frac{|x|^{2n+3}}{(2n+3)!}\bigg/\frac{|x|^{2n+1}}{(2n+1)!}= \frac{|x|^{2n+3}}{(2n+3)!}\frac{(2n+1)!}{|x|^{2n+1}}= \frac{|x|^2}{(2n+3)(2n+2)} $$ Since the limit of this as $n\to\infty$ is zero, we know that the series converges for all $x$.