Quick Question of convergence/divergence of an endpoint of an interval

1. Apr 26, 2012

anniecvc

1. The problem statement, all variables and given/known data
I have found the interval of convergence for the series
Ʃ (xn)/(3nn2) to be -3>x>3 and have tested +3 to break down to p series 1/n2 which converges.

However I am unsure of the negative endpoint of 3.
If I plug it in, the series looks like Ʃ(-3)n/(3nn2).
By alternating series test, the series should converge, but that is when the alternation is between -1 and 1. Here I have -3, which will alternate between larger and larger factors of 3 and -3. The denominator is 3n which given alone with the numerator would diverge, even though the denominator is multiplied by n2, I feel it is not sufficiently large enough for the limit to go to 0 as n -> ∞, since an exponential function grows more quickly than a quadratic.

I want to say it diverges, but don't have solid proof. Thoughts?

Last edited: Apr 26, 2012
2. Apr 26, 2012

LCKurtz

That simplifies to$$\sum \frac {(-1)^n}{n^2}$$doesn't it? What does the alternating series test say about this?

3. Apr 26, 2012

Dick

(-3)^n/3^n=(-1)^n. I'm really not sure what you are worried about.