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I'm wanting to find the radius of convergence and interval of convergence for a power series that isn't in the form $$\sum_{n=1}^{\infty} a_{n}x^{n}\tag{1}$$ but is instead in the form $$\sum_{n=1}^{\infty} a_{n}x^{2n+1}.\tag{2}$$ I'm having difficulty wrapping my head around this. The theorem I must work with gives the radius of convergence as $$R=\frac{1}{\beta}$$ where $\beta=\limsup_{n\to\infty}|a_{n}|^{1/n}$. One of the assumptions of this theorem though is that the power series looks like $(1)$. How would one generally massage $(2)$ to look like $(1)$ so that we could easily work with it?

Thanks in advance for your time, and apologies for the entry-level question.

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  • $\begingroup$ Hint: rename $a_n=b_{2n+1}$ and let $b_m=0$ for even $m$. Then you get an ordinary power series. $\endgroup$ – Wojowu May 9 '16 at 10:05
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    $\begingroup$ You can also use the substitution $y=x^2$ and find the $y$ radius of convergence. $\endgroup$ – Giuseppe Negro May 9 '16 at 10:06
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Hint:

$$\sum_{n=1}^{\infty} a_{n}x^{2n+1}\tag{2}=x\sum_{n=1}^{\infty} a_{n}(x^{2})^n$$ Substitute $u=x^2$. Find radius of convergence for u and then transform this back to $x$.

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  • $\begingroup$ I am curious, though, how does the $x$ you've pulled out the front above affect the series when you transform back? If I had a power series that was originally $\sum a_{n}x^{n+1}=x\sum a_{n}x^{n}$, would this have the same radius of convergence as $\sum a_{n}x^{n}$? $\endgroup$ – Old mate May 9 '16 at 10:23
  • $\begingroup$ As $x$ has infinite radius of convergence. $x\sum a_nx^n$ has the radius of the taylor series, as $r=\min(r_x,r_{sum})$.. $\endgroup$ – MrYouMath May 9 '16 at 10:40
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As Wojowu suggests, you can define $$b_m = \left\{ \begin{array}{l l} 0,& m = 2n\\ a_n,& m = 2n + 1 \end{array}\right.$$

For radius of convergence you need to calculate $\limsup_n\sqrt[n]{|b_n|}$ and the limit points are given by $\limsup_n\sqrt[2n]{|b_{2n}|} = 0$ and $\limsup_n\sqrt[2n+1]{|a_n|}\geq 0$, so the "even subsequence" can safely be ignored.

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  • $\begingroup$ @Old mate, you are welcome. $\endgroup$ – Ennar May 9 '16 at 10:21

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