# Help with radius of convergence of a power series.

I need to determine the radius of convergence of the series $\sum_{n=1}^\infty a_nx^n$, where $a_n=a^n+b^n$ and $a,b$ are real numbers. Not sure how to approach this one.

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By Cauchy's-Hadamard formula, with $\;R:=$ convergence radius, with the usual conventions when $\;R=0\,,\,\infty\;$ , we get:

$$\frac1R=\lim_{n\to\infty}\sup\sqrt[n]{|a^n+b^n|}$$

and assuming $\;|a|\ge|b|\;$ , we get

$$\sqrt[n]{|a^n+b^n|}=|a|\sqrt[n]{1+\left(\frac{|b|}{|a|}\right)^n}\xrightarrow[n\to\infty]{}|a|$$

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Is there a way to do this using the ratio test? – James Apr 19 '14 at 16:17
Yes, @James: $$\frac{|a^{n+1}+b^{n+1}|}{|a^n+b^n}|=|a|\frac{\left|1+\left(\frac ba\right)^{n+1}\right|}{\left|1+\left(\frac ba\right)^n\right|}\xrightarrow[n\to\infty]{}|a|$$ But C-H formula is easier in this case, imo. – DonAntonio Apr 19 '14 at 16:22

Since $a_n = a^n + b^n$, then

$$\sum_{n=1}^\infty a_n x^n =\sum_{n=1}^\infty (ax)^n + \sum_{n=1}^\infty (bx)^n .$$

For this to be convergent, both series must be convergent, but these are regular geometric progressions, so the conditions for their convergence are that $|ax|<1$, and $|bx|<1$. So we must have simultaneously

$$|x|< \frac{1}{|a|}$$

and

$$|x|<\frac{1}{|b|}$$

Which is equivalent to $$|x| < \frac{1}{\max(|a|,|b|)}.$$

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Simplest answer, I like it. – user88595 Apr 19 '14 at 16:21
Thank you for your help – James Apr 19 '14 at 16:29

Your series converges when $$\lim_{n \to \infty}\left\vert\sqrt[n]{\left \vert a^n+b^n\right \vert} x \right\vert < 1$$ Hence, the radius of convergence is $$R = \dfrac1{\lim_{n \to \infty} \sqrt[n]{\left \vert a^n+b^n\right \vert}} = \dfrac1{\max(\vert a \vert, \vert b \vert)}$$

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