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Show that the $\lim \inf a_n = \lim \sup a_n$ if and only if $\lim a_n$ exists.

Attempt at proof.

($ \Rightarrow $) Assume $\lim \inf a_n = \lim \sup a_n= L$. Then we have $|\sup a_n -L| \lt \epsilon$ and $|\inf a_n - L| \lt \epsilon$ and thus $$L-\epsilon \lt \sup a_n \lt \epsilon +L $$ and $$L -\epsilon \lt \inf a_n \lt L + \epsilon$$
and since $\inf a_n \le a_n\le \sup a_n$ it follows that $L- \epsilon \lt a_n\lt L + \epsilon$ i.e. $\lim a_n =L$

($\Leftarrow$)

Let $\lim a_n = L$. Assume that $\lim \inf a_n \neq \lim \sup a_n$ Choose $N_0$ such that $n \ge N_0$ implies $|a_n- L| \lt \epsilon = \frac{\sup a_n- \inf a_n}{2}$

Then we see that $$|\sup a_n - a_n| \le |a_n- L| \lt \frac{\sup a_n- \inf a_n}{2}$$

and $$|\inf a_n - a_n| \le |a_n- L| \lt \frac{\sup a_n- \inf a_n}{2}$$

By the Triangle Inequality we have $$|\sup a_n- \inf a_n| \lt |\sup a_n - a_n| + |a_n - \inf a_n| \lt 2\frac {\sup a_n- \inf a_n}{2}$$ which is a contradiction.

Can you please provide the correct proof. I'm pretty sure the second direction is incorrect. I am having a hard time showing that $|\sup a_n - a_n| \le |a_n- L|$ and that $|\inf a_n - a_n| \le |a_n- L|$ could be true. Thanks in advance.

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You have issues with both the directions and you need to understand the meaning of $\liminf, \limsup$ properly. The problem in first part is the inequality $\liminf a_n\leq a_n\leq \limsup a_n$ which is false (check using the sequence $a_n=(-1)^n(1+n^{-1})$).

And you have already mentioned the problem with second part. So let us now see how to do this properly. Assume $\liminf a_n=\limsup a_n=L$ and let $\epsilon >0$ be arbitrary. By definition of $\limsup, \liminf $ there exist positive integers $n_1,n_2$ such that $$L-\epsilon=\liminf a_n-\epsilon <a_n, \forall n\geq n_1\\ a_n<\limsup a_n+\epsilon=L+\epsilon, \forall n\geq n_2$$ and therefore if $n\geq n_0=\max(n_1,n_2)$ then we have $$L-\epsilon<a_n<L+\epsilon, \forall n\geq n_0$$ and this means that $\lim_{n\to\infty} a_n=L$.

Next the second direction is easy. Assume that $\lim_{n\to\infty} a_n=L$ and then we show that $\limsup a_n=\liminf a_n=L$. Let $\epsilon >0$ be arbitrary. Then there is a positive integer $n_0$ such that $$L-\epsilon<a_n<L+\epsilon,n\geq n_0$$ By definition of $\liminf, \limsup$ the conclusion follows from the above inequality.

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  • $\begingroup$ Thanks for taking the time out for proper answer! Appreciate it. $\endgroup$ – Red Jul 28 '18 at 4:01
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Your first implication is correct, but I don't think your second is. Here is how to do this, you can fill in the details.

$\sup a_n =\sup_{k\geq n} a_k$ here. $\inf a_n$ is analogous. That's just so the notation is simplified.

Let's consider $\epsilon>0$. There exist an $n$, so that for $k\geq n$: $$L-\epsilon<a_k<L+\epsilon $$ Taking supremum of such $k$, we get that: $$L-\epsilon\leq\sup a_n\leq L+\epsilon $$ This means that: $$\lim_{n\to \infty} \sup a_n = L $$ We can do the exact same reasoning with infimum.

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