# Alternating Series Test of Convergence

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}.

Is that correct?

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

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

$$\sum\limits_{n=1}^{\infty}\frac{(-5)^n(x-2)^n}{n}=\sum\limits_{n=1}^{\infty}\frac{(-5x+10)^n}{n}\tag1$$

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

$$\frac{9}{5}

The value of the sum is:

$$\sum\limits_{n=1}^{\infty}\frac{(-5)^n(x-2)^n}{n}=-\ln(1-|10-5x|)\tag4$$