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

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

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 $$-\infty \lt \liminf_{n\to \infty} a_n = \limsup_{n\to \infty} a_n \lt \infty$$

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.

Hint:

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