Is This Series Divergent? Geometric Series Manipulation and Long Term Behavior

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SUMMARY

The series in question is \sum\limits_{n=1}^\infty \frac{(-5)^{2n}}{n^{2}9^{n}}, which can be manipulated into \sum\limits_{n=1}^\infty \frac{1}{n^{2}}(\frac{25}{9})^{n}. The conclusion drawn by the student was that the series is divergent due to the geometric series component with r > 1. However, the presence of the \frac{1}{n^2} factor necessitates a deeper analysis of the series' long-term behavior, specifically \lim_{n\rightarrow \infty} {\frac{(\frac{25}{9})^n}{n^2}} = \infty, indicating divergence. The discussion highlights the importance of considering all components of a series when determining convergence or divergence.

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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?
 
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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.
 
HS-Scientist said:
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.
 

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