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I'm learning how to test infinite series in Calc II. I have a problem that says to use the ratio test to determine if a series converges.

The series is: $\sum\limits_{k=1}^{\infty}\dfrac{k^2}{4^k}$ so when I applied the ratio test, I got: $$\lim\limits_{k\to\infty}\dfrac{(k+1)^2}{4^{k+1}}\cdot\dfrac{4^k}{k^2}$$ Simplified to: $$\lim\limits_{k\to\infty}\dfrac{(k+1)^{2k}}{k^{2k+2}}$$


This would produce the indeterminate form of $\dfrac{\infty}{\infty}$ so I was going to use L'Hospital's rule and take the derivative of the top and bototm, but I'm having trouble with that - the deriviate of the numerator produces a very complex results and I don't think the problem's supposed to be this hard so I think I missed something in an earlier step. Can anyone point out where my error is?

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    $\begingroup$ Ratio Test applied wrongly. The ratio is $\frac{(k+1)^2}{k^2}\frac{1}{4}$. $\endgroup$ – André Nicolas Apr 2 '14 at 0:39
  • $\begingroup$ This is a polylogarithm. $\endgroup$ – Lucian Apr 2 '14 at 1:51
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The series is: $\sum\limits_{k=1}^{\infty}\dfrac{k^2}{4^k}$ so when I applied the ratio test, I got: $$\lim\limits_{k\to\infty}\dfrac{(k+1)^2}{4^{k+1}}\cdot\dfrac{4^k}{k^2}$$

Good. Here's where you go wrong:

Simplified to: $$\lim\limits_{k\to\infty}\dfrac{(k+1)^{2k}}{k^{2k+2}}$$

It actually simplifies to $$\lim\limits_{k\to\infty}\dfrac{1}4=\dfrac{1}4$$ Because $\lim\limits_{k\to\infty}\frac{(k+1)^2}{k^2}=1$ and $\lim\limits_{k\to\infty}\frac{4^k}{4^{k+1}}=\frac{1}{4}$

Here you can directly use the ratio test.

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You have the right idea, but you made a mistake in simplifying. Cheers!

$$\lim\limits_{k\to\infty}\dfrac{(k+1)^2}{4^{k+1}}\cdot\dfrac{4^k}{k^2} = \lim\limits_{k\to\infty}\dfrac{(k+1)^2}{4k^2}$$

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Hint: $$\lim_{k\to\infty}\dfrac{(k+1)^2}{4^{k+1}}\dfrac{4^k}{k^2}=\lim_{k\to\infty}\dfrac{(k+1)^2}{4k^2}\\ \implies \lim_{k\to\infty}\dfrac{(k+1)^2}{4k^2}=\lim_{k\to\infty}\dfrac{2(k+1)}{8k}=\dfrac{1}{4}<1$$

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