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I was wondering how would I prove that the sequence of {$\frac {n^2}{2^n}$} converges or diverges.

Normally, here I would set $\frac {n^2}{2^n}$ < epsilon and solve for n. However, this approach doesn't really work because I don't think I can isolate n. My next thought was to somehow use the squeeze lemma. The problem is I don't exactly know what to be squeezing the sequence between and even then I don't know if that is the correct approach. Any thoughts?

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    Have you already studied series? – 2017-02-23
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    @DonAntonio Yes, I was introduced to it very recently. Are you implying that the ratio test for sequences should be applied here? – 2017-02-23
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    That's an idea...that can be very good, indeed. – 2017-02-23

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Apply the ratio test:

$$\frac{(n+1)^2}{2^{n+1}}\frac{2^n}{n^2}=\frac12\left(\frac{n+1}n\right)^2\xrightarrow[n\to\infty]{}\frac12$$

and thus the series $\;\sum\limits_{n=1}^\infty\cfrac{n^2}{2^n}\;$ converges, and thus...

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    Ah okay I got it. Since the converges, the limit of the nth term must be zero.--) – 2017-02-23
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    @ΘΣΦGenSan Exactly so. ;) – 2017-02-23
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Let $a_n:=\frac{n^2}{2^n}$. Then $a_n^{1/n} \to 1/2<1$. Hence $\sum a_n$ converges. Therefore $a_n \to 0$