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 :)

• The series is written like $\sum a_{n}x^{n}$. You just need o identify your $a_n$'s. The ratio test is no good here because $a_k=0\lor a_{k+1}=0$. – Git Gud Jun 15 '14 at 19:17
• The series converges if $\lim_{n\to\infty}\left|\frac{x^{2n+3}(-3)^n}{x^{2n+1}(-3)^{n+1}}\right|\lt 1$, and diverges if the same limit is $\gt 1$. – André Nicolas Jun 15 '14 at 19:21

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 $x$.

• @mvfs314 For convergence you must have $\frac{|x|^2}{3}<1$, that is $|x|^2<3$ or $-\sqrt{3}<x<\sqrt{3}$. – egreg Jun 15 '14 at 19:39

Let

$$u_n(z)=\frac{x^{2n+1}}{(-3)^n}$$ then by the ratio test we have

$$\frac{|u_{n+1}(z)|}{|u_n(z)|}=\frac{|x|^2}{3}\xrightarrow{n\to\infty}\frac{|x|^2}{3}<1\iff|x|<\sqrt3$$ hence the radius of convergence is $R=\sqrt3$.

• Thanks, but I'm trying to use the power series tests (to get its radius). – mvfs314 Jun 15 '14 at 19:37

It's better to use this:

$\rho=\limsup_{n \to \infty} \sqrt[n]{|a_{n}|}$, then the series converges for all $x$ that $|x| < \frac{1}{\rho}$ and doesn't coverge for $|x|>\frac{1}{\rho}$. In this case:

$\limsup_{n \to \infty} \sqrt[n]{|a_n|}=\limsup_{n \to \infty} \sqrt[2n+1]{\frac{1}{3}}=\frac{1}{\sqrt{3}}$, so the series converges for all $x$ that $|x| < \sqrt{3}$ and doesn't coverge for $|x|>\sqrt{3}$. For $|x|=\sqrt{3}$ the series doesn't coverge, because $a_n=- \sqrt{3}$ or $a_n=\sqrt{3}$ for all $n$.

Since $x\ne0$, we can apply the ratio test for Absolute Convergence, which is stated precisely as so:

Let $\sum u_{k}$ be a series with nonzero terms and suppose that $$p= {Lim_\xrightarrow{k\to\infty}}{\frac{|u_k+1|}{|u_k|}}$$ (a) The series converges absolutely if $p<1$.

(b) The series diverges if $p>1$ or $p = \infty$

(c) The test is inconclusive if $p=1$

Applied in this scenario, we get: $$\frac{|x^{2(n+1)+1}|}{|-3^{n+1}|}* \frac{|-3^n|}{|x^{2n+1}|} =\frac{|x^2|}{|-3|}=\frac{x^2}{3}$$

Now $${Lim_\xrightarrow{n\to\infty}}{\frac{x^2}{-3}}<1\iff|x|<\sqrt3$$

Our radius of convergence is $\sqrt3$ and the interval of convergence is $-\sqrt3< x < \sqrt3$