# Find the interval of convergence of a power series with a single endpoint

The given series is: $$\sum^\infty_{n=0}\left(x^{2n+1}+2x^{2n+2}\right)$$ I know that the interval of convergence can be found by the ratio test, so what I tried was: $$\lim_{n \to \infty}\frac{x^{2n+3}+2x^{2n+4}}{x^{2n+1}+2x^{2n+2}}=x^2$$ Now we know that $$-1<|x^2|<1$$ or just $$x<1$$.

Then solving for the single endpoint: $$\sum^\infty_{n=0}\left(1^{2n+1}+2^{2n+2}\right)$$ $$\lim_{n\to\infty}1^{2n+1}+2^{2n+2}=\infty$$ So the endpoint diverges, and the interval is $$(-\infty,1)$$

• $|x|<1$ implies $-1<x<1$. You had your inequalities wrong. Apr 27, 2020 at 12:15
• The interval of convergence has to be centered in $0$, I can't see how could it be otherwise.
– Axel
Apr 27, 2020 at 12:16

The series does not converge for $$x \leq -1$$. A series $$\sum a_n$$ cannot converge unless $$a_n \to 0$$. Can you check that in this case the general term does not tend to $$0$$ if $$x \leq -1$$? The correct interval of convergence is $$(-1,1)$$.
• Why does the inequality change from $x^2$ to $x$? Apr 27, 2020 at 12:10
• @ChetBarkley I am not using ratio test at all. Ratio test for the entire series is is not too convenient here. Note that convergnce for $|x|<1$ can be derived by just writing the given series as sum of two geometric series. Apr 27, 2020 at 12:13