1. ## Intervals of Convergence

Find the intervals of convergence of f(x), f'(x) f''(x) and $\int {f(x)dx}$ for the function $f(x) = \sum_{n=1}^{\infty} ((-1)^{(n+1)} (x-2)^n) / n$ Check for convergence at endpoints of each inverval.

I've been having a lot of trouble with problem and I'd really appreciate any help I can get.

2. For f(x), apply the ratio test and find the limit as $n\rightarrow{\infty}$. Then check the endpoints as well. It converges at x=2.

By the ratio test:

$\frac{(-1)^{n+2}(x-2)^{n+1}}{n+1}\cdot\frac{n}{(-1)^{n+1}(x-2)^{n}}=\frac{n(2-x)}{n+1}$

$\lim_{n\rightarrow{\infty}}\frac{n(2-x)}{n+1}=|2-x|$

The ratio test gives cionvergence for $|2-x|<1$ and divergence for $|2-x|>1$

Solving the first inequality we get $1.

At x=1, the series is $-\sum_{n=1}^{\infty}\frac{1}{n}$, which is a divergent p-series.

At x=3, it is convergent and converges to -ln(2).

So, the convergence set is $1

Check it out. Make sure I did not err somewhere, then run a test on the derivatives of the series.

3. Originally Posted by galactus
At x=3, it is convergent and converges to -ln(2).
What test did you use to find that?

Check it out. Make sure I did not err somewhere, then run a test on the derivatives of the series.
How would you approach the other three?

4. Also, does anybody know how to integrate f(x)? We tried on a TI-89 but even that didn't give a definite answer.

5. You can use the alternating series test to show convergence.

But if x=3, you get $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..............$

This is a series I knew converges. This is the series associated with

the "Euler's constant". Which shows $1=ln(2)+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-.............$

As for the antiderivative of your series. Integrating wrt x, wouldn't it be

$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-2)^{n+1}}{n(n+1)}$?.