0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Yep, that's it. Looked at this problem for two hours and couldn't figure out this simple approach lol. $\endgroup$ – millionmilesaway Nov 27 '20 at 21:00
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.