How can I prove $\lim_{n \to \infty} \frac{n^2}{2^n}=0$ [duplicate]

Notice that this can be seen to be equivalent to proving that the following converges:

$$\sum_{k=1}^\infty a_k,\quad a_n=\frac{n^2}{2^n}$$

By the ratio test, we need to show that

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$$

but we can see that

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{2n^2}$$

which is where you were at. Now notice that

$$\frac{(n+1)^2}{2n^2}=\frac12\left(1+\frac1n\right)^2\stackrel{n\to\infty}\longrightarrow\frac12<1$$

One easy way is to use the binomial theorem to write for $n\ge 3$

$$2^n=(1+1)^n=\sum_{k=1}^n\binom{n}{k}\ge \frac{1}{6}n(n-1)(n-2)$$

Then we have

$$\frac{n^2}{2^n}\le \frac{6}{(n-1)(1-2/n)}$$

Can you finish now?

Another way is to note that $2^n=e^{n\log(2)}\ge \frac16 (\log(2)n)^3$ so that

$$\frac{n^2}{2^n}\le \frac{6}{\log^3(2) n}$$

A third way is to observe that

$$\lim_{n\to \infty}\left(\frac{n^2}{2^n}\right)^{1/n}=\frac12$$

Thus, for any $\epsilon>0$ there exists a number $N$ such that

$$\left(\frac{n^2}{2^n}\right)\le \left(\frac{1}{2}+\epsilon\right)^n$$

whenever $n>N$. Take $\epsilon=1/4$. Then, there exists a number $N$ such that

$$\left(\frac{n^2}{2^n}\right)\le \left(\frac34\right)^n$$

whenever $n>N$.

If $a_n=\frac{n^2}{2^n}$ then

$$ \frac{a_{n+1}}{a_n}=\frac{(1+1/n)^2}{2}$$

For $n>2$, show $$\frac{(1+1/n)^2}{2}\leq \frac{8}{9}$$

So $$a_{n+3}\leq \left(\frac{8}{9}\right)^n a_3$$