Convergence of the series at x = -3 endpoint

[anniecvc]

Homework Statement

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?

 

[LCKurtz]

Homework Statement

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?

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

 

[Dick]

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