# Find the radius of convergence and interval of covergence of the series.

I'm in the process of trying to learn about the topic of series, power series, etc. (when I say learn I mean self-teach... everything urgh). My general understanding of the topic of Series as a whole is quite low, so a little more insight would be appreciated. Either I don't really take in much from reading or my book does an awful job of explaining it, however...

We have... Find the radius of convergence and interval of convergence of the series...

$\sum^\infty_{n=1}(-1)^nnx^n$

So to do this I try the ratio test (since they used this in the book for most of the examples)...

$|\frac{a_{n+1}}{a_n}| = |\frac{(-1)^{n+1}(n+1)x^n}{(-1)^n(n)x^n}| = |\frac{-1(n+1)}{n}| = |\frac{-n-1}{n}|= |-1 -\frac{1}{n}|$

Hence, this tends towards $|-1|$ or simply $1$ as $n \rightarrow \infty$

Ratio test says the series converges if 1 < 1 or diverges if 1 > 1. Since 1 = 1 this means the ratio test doesnt work right?

It looks like it is an "Alternating Series" but I again, I have no idea how to apply this...

Can anyone help me out... this is supposedly a basic question but nothing's clicking for me.

• $a_{n+1} = (-1)^{n+1}(n+1)x^{n+1}$, you wrote $x^n$. There's a factor $x$ in the quotient, and you get the limit $\lvert x\rvert$. Mar 26 '14 at 14:51

$$\left|\frac{A_{n+1}}{A_n}\right|=\left|\frac{(n+1)x^{n+1}}{nx^n}\right|=\frac{n+1}n|x|\xrightarrow[n\to\infty]{}|x|$$
so the series converges absolutely for $\;|x|<1\;$ ...
For the radius of convergence apply the Cauchy-Hadamard formula...which gives $\;1\;$ , of course.