Determine the convergence of $ \sum_{n=2}^ \infty \frac{(1)^n}{n(n-1)}$ and $ \sum_{n=2}^ \infty \frac{(-1)^n}{n(n-1)}$

Question

I'm trying to analyze the ratio of convergence of the following power series, I also want to know if the series converges at the endpoints of R (radius of convergence).

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

By the ratio test, it is easy to see that$\ R=1$, that is, the series converges for$\ |x|<1$

I want to know what happens with the series at$\ x=1$ and$\ x=-1$

At$\ x=1$

$$ \sum_{n=2}^ \infty \frac{(1)^n}{n(n-1)}$$

I'm struggling with this problem.

Answer 1

Since $n(n-1) \ge \frac12 n^2$ if $n\ge 2$, we have $$ \sum_{n=2}^ \infty \frac{(-1)^n}{n(n-1)} \le \sum_{n=2}^ \infty \frac{(1)^n}{n(n-1)} \le \sum_{n=2}^ \infty \frac{2}{n^2} \le 2\sum_{n=1}^ \infty \frac{1}{n^2} = \frac{\pi^2}{3} $$ Therefore, both series converge.

Answer 2

If you use partial fraction decomposition on $\frac{1}{n(n-1)}$, you get $\frac{1}{n-1} - \frac{1}{n}$. So the sequence will eliminate all terms except $\frac{1}{2-1}$, thus, the sum is equal to 1.

Answer 3

"By the ratio test, it is easy to see that$\ R=1$, that is, the series converges for$\ |x|<1$" You should add "and diverges for $|x|>1.$" At $x=\pm 1$ the series of absolute values is $\sum_{n=2}^ \infty \frac{1}{n(n-1)}.$ The latter series converges by the limit comparison test, when compared with $\sum_{n=2}^ \infty \frac{1}{n^2}<\infty.$ It follows that the given series converges, absolutely, on $[-1,1].$