I am trying to find the radius of convergence and interval of convergence for the sum $$ \sum_{n=1}^{\infty}\frac{(-5)^n(x-2)^n}{n} $$ I recognize this is an alternating series. I do not know how to go about solving for the radius and interval of convergence though. Is it simply to find.. $$ \lim_{n\rightarrow\infty}\lvert\frac{(5)^{n+1}(x-2)^{n+1}}{n+1}*\frac{n}{5^n(x-2)^n}\rvert=\lim_{n\rightarrow\infty}\lvert5(x-2)\rvert\frac{n}{n+1} $$ This then implies that $\lvert(x-2)\rvert<\frac{1}{5}$ is the radius of convergence and the interval of convergence would be $\frac{9}{5}<x<\frac{11}{5}$.

Is that correct?

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    $\begingroup$ That is correct. But you must also check for convergence at the two endpoints of the interval. $\endgroup$ – John Wayland Bales Apr 4 at 17:26
  • $\begingroup$ what does that entail? $\endgroup$ – joseph Apr 4 at 17:51
  • $\begingroup$ @JohnWaylandBales When $x\lt2$, that is not an alternating series. A different test must be used. $\endgroup$ – FredH Apr 4 at 19:02
  • $\begingroup$ @FredH The series alternates regardless of the sign of $x-2$. $\endgroup$ – John Wayland Bales Apr 4 at 19:07
  • $\begingroup$ @JohnWaylandBales If $x = 1.9$, all terms of the series are positive. How is that alternating? $\endgroup$ – FredH Apr 4 at 19:09

Other way to determine the radius of the convergence is to calculate the value of the sum:


Let $t=10-5x$ $\hspace{0.5cm}$ and we can realize that $\int t^{n-1}dt=\frac{t^n}{n}$ $\hspace{0.5cm}$ we get:

$\sum\limits_{n=1}^\infty \int t^{n-1}dt=\int \sum\limits_{n=1}^\infty t^{n-1}dt\tag2$

The geometrical series convergent if $|t|\lt 1$ that is $|10-5x|\lt 1$ so the series is convergent if


The value of the sum is:



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