Find the limit of the sequence $a_n=\frac {cos^2n}{2^n}$

Since $0\le\frac {cos^2n}{2^n}\le\frac{1}{2^n},$ for all $n$

The Squeeze Theorem:

if $f(n)\le g(n) \le h(n)$ when $n$ is near $a$ and $\lim\limits_{n \to a}f(n)=\lim\limits_{n \to a}h(n)=L$ then $\lim\limits_{n \to a}g(n)=L$

Here, $f(n)=0,g(n)=\frac {cos^2n}{2^n}$ and $h(n)=\frac{1}{2^n}$

What I don't understand: How were the upper and lower functions of the inequality $f(n)\le g(n) \le h(n)$ chosen?

$\lim\limits_{n \to \infty}\frac{1}{2^n}$ and $\lim\limits_{n \to \infty}0$ are simple enough to evaluate. But how is it that the author was able to pick such simple upper and lower functions to evaluate?

Any help is greatly appreciated

  • 1
    $\begingroup$ A certain amount of experience ist necessary. And this experience comes when doing examples etc. $\endgroup$ Mar 7, 2020 at 8:01
  • $\begingroup$ First, examine the function. Where do you think it is headed as the input grows? Plug in some numbers, graph it. Once you have some idea, then you start looking for simpler, smaller- and larger-valued functions which have the same limit. $\sin$ and $\cos$ and their powers/ products always have the nice property that they;re trapped between $-1$ and $1$, while even powers are trapped between $0$ and $1$. For example, $-1\le \sin^2(2n+5)\cdot \cos^7(n^3)\le 1$ where $0\le \sin^8(5n)\cos^2(n/3)\le 1$ $\endgroup$
    – David P
    Mar 7, 2020 at 8:18

2 Answers 2


You are able to pick any function that you know is bigger or smaller then $g(n)$.

Another example for your function $g(n) = \frac{\cos^2{n}}{2^n}$ would be:

$$\frac{\cos^2{n}}{2^{n+1}} \leq \frac{\cos^2{n}}{2^n} \leq \frac{\cos^2{n}}{2^{n-1}}$$

Now, not every choice in $f(n)$ and $h(n)$ will actually be helpful. You want to chose them in a way that makes it easy to calculate.

Hope it helps.


Well the aim is to go from something you don't know a priori the behaviour to something you are able to deal with.

So any rough inequality will do, and I would say the simpler the better.

In our case $\cos$ is a bounded function of $[-1,1]$ so $0\le \cos^2 n\le 1$, and we have gone from something which depends of $n$ to constants, this is far better.

Finally the author gets to $0\le a_n\le \dfrac 1{2^n}$, but you could argue, that it is still too complicated and could say that $2^n > n$ so that $0\le a_n\le \dfrac 1n$ and now you can apply the squeeze theorem with an even simpler statement.

You have big freedom in your choices when reducing inequalities.

For instance lets take $b_n = \dfrac{n^2+1}{n^3+2^n}$ still obviously converging to $0$.

  • first try could be $n^2+1<2n^2\quad$ and $\quad n^3+2^n>2^n$

so you get $\quad 0\le b_n\le\dfrac{2n^2}{2^n}\quad$ you can now conclude saying $2^n\gg n^2$.

  • but what about the very basic $2^n>n$ and then $0\le b_n\le \dfrac{n^2+1}{n^3+n}=\dfrac 1n$

You see that the inequality we have chosen is not the most precise, we have now something that converges slowly to $0$ compared to $\dfrac{n^2}{2^n}$ yet it is quite enough to conclude.

You will say, you cheated it works because there was $n^2+1$ on numerator and it simplified, what if there was $n^2+3$ instead?

Well then just go $3<n^2$ and $2^n>0$ (at least for $n$ large enough) so that $\dfrac{n^2+3}{n^3+2^n}\le \dfrac{2n^2}{n^3}=\dfrac 2n$ and you still can conclude.

You see that we madly replaced the most contributing term $2^n$ by just $0$, yet it does not change our conclusion.

In these matters, you have to go straight for maximum simplification, if you still can conclude then it was adequate. If you reach an undetermined limit then and only then go for more precise inequalities...


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