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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 $(-\infty,1)$

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

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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 $(-1,1)$.

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  • $\begingroup$ Why does the inequality change from $x^2$ to $x$? $\endgroup$ Apr 27, 2020 at 12:10
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    $\begingroup$ @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. $\endgroup$ Apr 27, 2020 at 12:13

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