3

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

I tried to use $\mid \frac{n^2}{2^n} - 0 \mid <\epsilon $ However, because of $n^2$ I cannot use it. Also, I tried to use ratio test and I got $\lim_{n->\infty}\frac{(n+1)^2}{2}\frac{1}{n^2}$, but after that one, I dont know how to get limit < 0

I have no idea to solve this problem.

I can also use a squeez lemma, but dont know how to apply to this question.

Mark Viola
  • 179,405
user421044
  • 97

5 Answers5

4

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

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

Simply Beautiful Art
  • 74,685
3

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 $n>N$.

Mark Viola
  • 179,405
  • but I don't think I can use binomial theorem to prove it. I have to use ratio test, definition of limit, or squeeze lemma only :/ – user421044 Mar 10 '17 at 00:08
  • @user421044 I've added two more ways forward. -Mark – Mark Viola Mar 10 '17 at 00:15
  • The last method may be changed to the root test ;) – Simply Beautiful Art Mar 10 '17 at 00:36
  • @SimplyBeautifulArt The mission is different. Here, all we need to show is that the sequence $a_n = n^2/2^n\to 0$ while for the series $\sum_{n=1}a_n$, we need to show that the series converges. The "root-test' as applied to the latter shows that if $a_n^{1/n}\to a<1$, then the series converges. Obviously, a corollary is that $a_n\to 0$ if $a_n^{1/n}\to a<1$. – Mark Viola Mar 10 '17 at 02:41
1

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

Thomas Andrews
  • 177,126
0

We know $$\ell=\lim_{n \to \infty} \dfrac{n}{2^n}=\lim_{n \to \infty} \dfrac{n+1}{2^{n+1}}=\frac12\lim_{n \to \infty} \dfrac{n}{2^n}+\lim_{n \to \infty} \dfrac{1}{2^{n+1}}=\frac12\ell+0=\frac12\ell$$ so $\ell=0$.

$$\ell=\lim_{n \to \infty} \frac{n^2}{2^n}=\lim_{n \to \infty} \frac{(n+1)^2}{2^{n+1}}=\frac12\lim_{n \to \infty} \frac{n^2}{2^n}+\lim_{n \to \infty} \frac{n}{2^n}+\lim_{n \to \infty} \frac{1}{2^{n+1}}=\frac12\ell+0+0$$ so $\ell=0$.

Nosrati
  • 29,995
0

Stolz-Ces$\mathrm{\grave{a}}$ro Theorem:

$$ \lim_{n \to \infty}{n \over 2^{n}} = \lim_{n \to \infty}{\left(n + 1\right) - n \over 2^{n + 1} - 2^{n}} = \lim_{n \to \infty}{1 \over 2^{n}} = \bbox[#ffe,10px,border:1px dotted navy]{\displaystyle{\large 0}} $$

Felix Marin
  • 89,464