Radius of convergence of series

Question

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. — colormegone, Apr 20 '14 at 04:07
You also could recognize the Taylor-Mc Laurin series of $\log(x)$ built at $x=1$. — Claude Leibovici, Apr 20 '14 at 06:01

score 1 · Accepted Answer · answered Apr 20 '14 at 04:07

1

The ratio test yields:

$$\lim\limits_{j \to \infty} \left|\dfrac{\frac{(-1)^j}{j+1}(x-1)^{j+1}}{\frac{(-1)^{j-1}}{j}(x-1)^j}\right| = \lim\limits_{j \to \infty} \frac{j}{j+1}|x-1| = |x-1|$$

So the series converges when $|x-1|<1$ and diverges when $|x-1|>1$.

Endpoints need to be checked separately. :)

answered Apr 20 '14 at 04:07

Bill Cook

29,244

Hmm, and how would I check the endpoints? – kiwifruit Apr 20 '14 at 04:12
Just plug them in: $x=0$ and $ x=2$. One gives a divergent harmonic series. The other is a convergent alternating harmonic series. – Bill Cook Apr 20 '14 at 22:05

Radius of convergence of series

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