Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

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

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

share|improve this answer
    
Is there a way to do this using the ratio test? –  James Apr 19 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 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 $$|x| < \frac{1}{\max(|a|,|b|)}.$$

share|improve this answer
    
Simplest answer, I like it. –  user88595 Apr 19 at 16:21
    
Thank you for your help –  James Apr 19 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 $$R = \dfrac1{\lim_{n \to \infty} \sqrt[n]{\left \vert a^n+b^n\right \vert}} = \dfrac1{\max(\vert a \vert, \vert b \vert)}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.