# Radius of convergence of series

Find the radius of convergence of this series:

$$f(x)= \sum_{j=1}^{\infty} \ \frac{(-1)^{j-1}}{j}(x-1)^j$$

I'm not sure what test to use to get the necessary result. I tried using the root test, but got an expression with both x and j that I can't infer from.

Edit: how can I check convergence at the endpoints? Is it just by plugging in values?

• The Root Test isn't applicable, since the denominator is just $\ j \$ , not the $\ j$ th power of something. Try the Ratio Test. Apr 20 '14 at 4:07
• You also could recognize the Taylor-Mc Laurin series of $\log(x)$ built at $x=1$. Apr 20 '14 at 6:01

$$\lim\limits_{j \to \infty} \left|\dfrac{\frac{(-1)^j}{j+1}(x-1)^{j+1}}{\frac{(-1)^{j-1}}{j}(x-1)^j}\right| = \lim\limits_{j \to \infty} \frac{j}{j+1}|x-1| = |x-1|$$
So the series converges when $|x-1|<1$ and diverges when $|x-1|>1$.
• Just plug them in: $x=0$ and $x=2$. One gives a divergent harmonic series. The other is a convergent alternating harmonic series. Apr 20 '14 at 22:05