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Find an appropriate power series and apply the ratio test to show the following infinite sum converges: $$ \sum_{n=0}^\infty \frac{n^2}{5^n} $$

I'd simply do it like this: $$ \text{Let } a_n = \frac{n^2}{5^n}. \text{Then by the ratio test we have}\\ \begin{align*} \left|{\frac{a_{n+1}}{a_n}}\right|= \left|\frac{(n+1)^2}{5^{n+1}}\cdot \frac{5^n}{n^2}\right| = \left|\frac{n^2+2n+1}{5n^2}\right| \end{align*}\\ \implies \lim_{n\to\infty} \frac{n^2+2n+1}{5n^2} = \lim_{n\to\infty} \frac{1+\frac{2}{n}+\frac{1}{n^2}}{5} = \frac{1}{5} < 1 $$

I'm confused since I thought I could directly apply the ratio test on this series. (without finding a power series)

I guess my professor wants me to solve it like that, but I have no idea what kind of convergent power series I could (upper-)bound this with.

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3 Answers 3

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Your approach is fine. The approach that the question might have been looking for is to consider the power series $\sum_{n=0}^\infty a_n x^n$ where $a_n=n^2$ and check that $x=1/5$ is within the radius of convergence.

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  • $\begingroup$ Yep, that's it. Looked at this problem for two hours and couldn't figure out this simple approach lol. $\endgroup$ Commented Nov 27, 2020 at 21:00
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Hint:

Work with $\sum\limits_{n=0}^\infty\ x^n\;,\;\;|x|<1\;$ , and then differentiatie twice...In fact, you should also find out what this series converges to.

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Hint

Consider the geometric power series $$\sum x^n$$ with $ R=1 $ as radius of convergence.

its derivatives at $ x\in(-1,0)\cup(0,1)$ $$\frac 1x\sum nx^n$$ and $$\sum n^2x^n$$ have the same Radius.

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