# Joanne 's question at Yahoo! Answers (Interval of convergence)

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello Joanne,

The series is $\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{(n-1)2^n}(x-1)^n.$ Using the ratio test: \begin{aligned}L&=\lim_{n\to \infty}\left|\frac{u_{n+1}}{u_n}\right|\\&=\lim_{n\to \infty}\left|\dfrac{(-1)^{n+1}(x-1)^{n+1}}{n2^{n+1}}\cdot\frac{(n-1)2^n}{(-1)^n(x-1)^n}\right|\\&=\lim_{n\to \infty}\left|\dfrac{n-1}{2n}(x-1)\right|\\&=\frac{|x-1|}{2}<1\\&\Leftrightarrow |x-1|<2\\& \Leftrightarrow x\in (-1,3)\end{aligned}

So, the series is absolutely convergent if $|x-1|<2$ and divergent if $|x-1|>2$. If $|x-1|=2$ (i.e. $x=-1$ or $x=3$) we have

$(a)\;x=-1$. The series is $\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{(n-1)2^n}(-2)^n=\displaystyle\sum_{n=2}^{\infty}\dfrac{1}{n-1}.$ Using the limit comparison test we easily verify that the series is divergent.

$(b)\;x=3$. The series is $\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{(n-1)2^n}2^n=\displaystyle\sum_{n=2}^{\infty}\dfrac{(-1)^n}{n-1}.$ Using the Lebniz alteranting series criterion we easily verify that the series is conditionally convergent.

As a consequence the given series is convergent iff $x\in(-1,3]$