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Radius of convergence of the power series $$\sum_{n=1}^\infty (2+(-1)^n)^nx^n$$ is $$1/3.$$ But I can't conclude that the power series is divergent at the end points​. How can I solve this problem?

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Step 1: Plug in $x=\frac{1}{3}$.

$\displaystyle \sum_{n=1}^\infty (2+(-1)^n)^n\left(\frac{1}{3}\right)^n=\sum_{n=1}^\infty \left(\frac{2}{3}+\frac{(-1)^n}{3}\right)^n$

This series is divergent because for every second term is $1$ and every other term is monotonically decreasing from the one two before it. So you can break into two series, one which converges $\frac{1}{3}+\frac{1}{27}+\frac{1}{243}...$ and one which diverges: $1+1+1...$

Step 2: Plug in $x=-\frac{1}{3}$

$\displaystyle \sum_{n=1}^\infty (2+(-1)^n)^n\left(-\frac{1}{3}\right)^n= \sum_{n=1}^\infty \left(-\frac{2}{3}-\frac{(-1)^n}{3}\right)^n$

This is similar, as every second term is $1$, and every other term alternates in a monotonically decreasing sequence from the second term before it. So you can break into two series, one which converges $-\frac{1}{3} - \frac{1}{27} - \frac{1}{243}...$ and one which diverges $1+1+1...$

Remember that the sum of a divergent and convergent series is divergent.

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  • $\begingroup$ (-1)^2n = 1 @Did $\endgroup$ – Saketh Malyala May 29 '17 at 19:38
  • $\begingroup$ Right. Sorry for the noise. $\endgroup$ – Did May 29 '17 at 19:44

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