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I need to calculate limit $$\lim\limits_{n \to \infty} n^2 q^n,$$ where $|q|<1$. Any hints how to do that would be appreciated.

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First function is a polynomial which goes to infinity, and the second one is an exponential which goes to zero. And exponential will overpower any polynomial eventually. Therefore, $$\lim\limits_{n \to \infty} n^2 q^n=0$$

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    $\begingroup$ You state a result "exponential will overpower any polynomial eventually" as if it is obvious. the problem here is actually trying to ask "why an exponential will overpower any polynomial eventually". And thus an answer requires a proof that this limit is $0$ by using standard limit rules and formulas. On the other hand it seems strange the answer has been accepted. $\endgroup$ – Paramanand Singh Jun 11 '14 at 3:51
  • $\begingroup$ @ParamanandSingh accepted answer does not mean that it is the best answer, it only means that it worked for the OP. I tried to give some sort of graphical intuition, because I knew other people will provide wonderful mathematical proof. You are right though, even I would not have accepted my own answer. $\endgroup$ – Edwin_R Jun 11 '14 at 7:36
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$$ \lim_{n\to\infty}n^2q^n=\lim_{n\to\infty}n^2e^{n\log\mid q\mid}, $$

Notice that $\mid q\mid\lt1$ then we know that $\log(\mid q\mid)$ is negative. Call $\log(\mid q\mid)=-\alpha$, where $\alpha\gt0$.

Hence,

$$ \lim_{n\to\infty}n^2q^n=\lim_{n\to\infty}n^2e^{-n\alpha}=0. $$

Because $x^pe^{-ax}\underbrace{\rightarrow}_{\text{as}\, x\, \text{goes to}\, \infty}0$ for all $a\gt0$ ("exponentials beat polynomials").

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Another fancy way: take the series

$$\sum_{n=1}^\infty n^2q^n\;,\;\;|q|<1$$

Applying the $\;n$-th root test we get:

$$\sqrt[n]{n^2|q|^n}=\sqrt[n]{n^2}\;|q|\xrightarrow[n\to\infty]{}|q|<1$$

and thus the series converges (even absolutely), and from here we get at once that

$$n^2q^n\xrightarrow[n\to\infty]{}0$$

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  • $\begingroup$ it would have been better to apply the ratio test. because that would not need the slightly difficult limit of $\sqrt[n]{n^{2}}\to 1$. but otherwise nice +1. $\endgroup$ – Paramanand Singh Jun 11 '14 at 3:49

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