# Limit superior and inferior

How can I find the limit superior/inferior of $a_n$, as $n \rightarrow \infty$?

$$a_n=\frac{n^2+4n-5}{n^2+9}\sin^2\left(\frac{n\pi}{4}\right), n \in \mathbb N$$

I've tried Wolfram|Alpha, but it does not support limits superior or inferior.

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I added just two parentheses. –  Babak S. Mar 24 '13 at 20:54
$\limsup$ is the greatest and $\liminf$ is the smallest congestion point of the sequence. –  Berci Mar 24 '13 at 20:56

## 3 Answers

$\frac{n^2+4n-5}{n^2+9}$ tends to 1, and the second term oscillates between $0$ and $1$.

So : the limit superior is $1$, and the limit inferior is $0$.

Complete proof :

$a_n \geqslant 0$, so $\liminf a_n \geqslant 0$. $a_n \leqslant \frac{n^2+4n-5}{n^2+9}$, which tends to $1$, so $\limsup a_n \leqslant 1$.

Now, $a_{4n} = 0$ so $\liminf a_n = 0$. And $a_{4n+2} = \frac{n^2+4n-5}{n^2+9}$, which tends to $1$. So $\limsup a_n = 1$.

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Find subsequences that tend to $1$, and to $0$. –  Berci Mar 24 '13 at 20:58
+1 Ah, so painfully simple...and Berci: that's for the OP to complete. –  DonAntonio Mar 24 '13 at 20:59
I was mistaken, I think. I made a correction. –  Arnaud Mar 24 '13 at 21:04
why is the second term equals to 1/2? i think that that limit doesn't exist, does it? –  Miloš Lukačka Mar 24 '13 at 21:11
@Arnaud , you seem to have been influenced to believe something was wrong: just take $$\,n=8k+2\Longrightarrow \sin^2\frac{(8k+2)\pi}{4}=\sin^2\left(4k\pi+\frac{\pi}{2}\right)=1\ldots$$ –  DonAntonio Mar 24 '13 at 21:15

$a_n$ is a product of two terms. One of them has an ordinary limit, so problem is in the second term. Can you find $\limsup\limits_{n\to{\infty}}{\sin^2\left(\frac{n\pi}{4}\right)}$ and $\liminf\limits_{n\to{\infty}}{\sin^2\left(\frac{n\pi}{4}\right)}?$

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I am OK with finding limits, I just have some problems with finding the supremum.. –  Miloš Lukačka Mar 24 '13 at 21:02
On the sequence $n_k^{(1)}=4k+2$ we have ${\sin^2\left(\frac{n_k^{(1)}\pi}{4}\right)}=1 \;\;\Rightarrow \sup{\sin^2\left(\frac{n\pi}{4}\right)}=1$ and on the sequence $n_k^{(2)}=4k:\;\;{\sin^2\left(\frac{n_k^{(2)}\pi}{4}\right)}=0 \;\;\Rightarrow \inf{\sin^2\left(\frac{n\pi}{4}\right)}=0.$ –  M. Strochyk Mar 24 '13 at 21:11

Here is a relevant result: If $\lim a_n = a >0$, then $\limsup_n a_n b_n = a \limsup b_n$, and $\liminf_n a_n b_n = a \liminf b_n$.

To see this, let $\overline{b} = \limsup_n b_n$.

If $\overline{b} = \infty$, then some subsequence $b_{n_k} \to \infty$, and hence $\limsup_n a_n b_n = \infty$.

If $\overline{b} = -\infty$, then for all $L>0$, there exists $N$ such that if $n \ge N$, then $b_n \le -L$. So, choose $N$ such that $b_n \le -\frac{2}{a}L$ for all $n \ge N$ and $a_n > \frac{1}{2} a$. Then $a_n b_n \le -a_n \frac{2}{a}L \le -L$. It follows that $\limsup_n a_n b_n = -\infty$.

Otherwise, suppose $\overline{b} \in \mathbb{R}$. By definition, for some subsequence we have $b_{n_k} \to \overline{b}$. Since $a_{n_k} b_{n_k} \to a\overline{b}$ along the same subsequence, we have $\limsup_n a_n b_n \ge a\overline{b}$. Furthermore, for any $\epsilon>0$, there exists an $N$ such that if $n \ge N$, then $b_k \le \overline{b}+\epsilon$. So, choose $N$ such that when $n \ge N$, we have $b_k \le \overline{b}+\frac{\epsilon}{3a}$ $|a_k-a| < \frac{\epsilon}{2(|\overline{b}|+1)}$ and $a_k \le \frac{3}{2}a$. Then we have $$a_k b_k \le a_k \overline{b}+a_k \frac{\epsilon}{3a} \le a\overline{b}+(a_k-a)\overline{b}+\frac{3}{2}a\frac{\epsilon}{3a} \le a\overline{b}+|a_k-a||\overline{b}|+ \frac{\epsilon}{2} < a\overline{b} + \epsilon$$ from which it follows that $\limsup_n a_n b_n \le a\overline{b}$, and hence equal.

The corresponding result for $\liminf$ follows because $\liminf_n a_n b_n = - \limsup_n -a_n b_n = - a \limsup_n (a_n(- b_n)) =- a\limsup_n - b_n = a \liminf_n b_n$.

In the example above, let $a_n=\frac{n^2+4n-5}{n^2+9}$, $b_n =\sin^2 (\frac{n\pi}{4} )$, from which we find $a_n \to 1$, and $b_n = (0, \frac{1}{2}, 1, \frac{1}{2}, 0,...)$. It should be clear that $\limsup_n b_n = 1$, $\liminf_n b_n = 0$, from which the desired result follows (ie, the $\liminf$ of the sequence in question is $0$, and the $\limsup$ is $1$).

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