Convergence of a real sequence iff $-\infty \lt \liminf_{x\to \infty} a_n = \limsup_{x\to \infty} a_n \lt \infty$

Question

Could someone help me with this exercise? How would I go about this, how would I start?

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Let $(a_n)_{n \in \mathbb N}$ be a real sequence. Show that:

$(a_n)_{n \in \mathbb N}$ is convergent in $\mathbb R$ if and only if $$-\infty \lt \liminf_{n\to \infty} a_n = \limsup_{n\to \infty} a_n \lt \infty$$

Answer 1

Start from the definitions:

$$\limsup_{n\rightarrow \infty} a_n:=\lim_{n\rightarrow \infty} (\sup_{m\geq n}a_m)$$ $$\liminf_{n\rightarrow \infty} a_n:=\lim_{n\rightarrow \infty} (\inf_{m\geq n}a_m)$$

$$\lim_{n\rightarrow \infty} (\inf_{m\geq n}a_m)\leq \lim_{n\rightarrow \infty} (\sup_{m\geq n}a_m)$$

Notice that $(\sup_{m\geq n}a_m)_{n\in\mathbb N}$ and $(\inf_{m\geq n}a_m)_{n\in\mathbb N}$ can be seen as subsequences of $(a_n)_{n\in \mathbb N}$.

And we know (as AnotherJohnDoe suggested in the comments) that all the subsequences of a convergent sequence must converge to the same limit (the limit of the original sequence).

So assuming that $\lim_{n\rightarrow \mathbb N} a_n$ exists (is not $\pm\infty$) we must have that $$\limsup_{n\rightarrow \infty}a_n=\liminf_{n\rightarrow \infty}a_n=\lim_{n\rightarrow \mathbb N} a_n$$

Conversely assume that

$$-\infty<\liminf_{n\rightarrow \infty}a_n=\limsup_{n\rightarrow \infty}a_n<\infty$$

$$-\infty<\lim_{n\rightarrow \infty} (\inf_{m\geq n}a_m)=\lim_{n\rightarrow \infty} (\sup_{m\geq n}a_m)<\infty$$

Because $(\inf_{m\geq n}a_m)\leq a_n\leq(\sup_{m\geq n}a_m)$ you just take limits and noticing that the extremes are equal then the sequence must converge.

Answer 2

Hint:

For every $\epsilon>0$ there exists some $k_0\in\Bbb N$ such that $$-\epsilon+\liminf a_n<a_k<\limsup a_n+\epsilon$$ for all $k\ge k_0$.