Does factoring out 1/n² make a divergent geometric series valid?

[iRaid]

Homework Statement

Well I have a series that I solved one way, but my professor solved another and I'm wondering if my way is ok.
$$\sum\limits_{n=1}^\infty \frac{(-5)^{2n}}{n^{2}9^{n}}$$

Homework Equations

The Attempt at a Solution

Alright well I started out by changing it to:
$$\sum\limits_{n=1}^\infty \frac{1}{n^{2}}(\frac{(-5)^{2n}}{9^{n}})=\sum\limits_{n=1}^\infty \frac{1}{n^{2}}(\frac{25}{9})^{n}$$

So I concluded since the second part is a geometric series with r>1, it's divergent.

Is this allowed?

 

[Infrared]

Your manipulations are legal, but how do you conclude that just because the "second part" of the series is divergent, that the entire series is divergent. There is, after all, a factor of $\frac{1}{n^2}$ that reduces each term. You could look at long term behavior of the terms however and note that $\displaystyle\lim_{n\rightarrow \infty} {\frac{(\frac{25}{9})^n}{n^2}} = \infty$.

 

[iRaid]

Your manipulations are legal, but how do you conclude that just because the "second part" of the series is divergent, that the entire series is divergent. There is, after all, a factor of $\frac{1}{n^2}$ that reduces each term. You could look at long term behavior of the terms however and note that $\displaystyle\lim_{n\rightarrow \infty} {\frac{(\frac{25}{9})^n}{n^2}} = \infty$.

Very interesting, never thought about doing that. Could of just done the standard ratio test, which I think would work out easier, but hey this is a good way to do this problem I think.

Thanks.