Convergence of Limsup and Liminf

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\sequence {x_n}$ be a sequence in $\R$.

Let the limit superior of $\sequence {x_n}$ be $\overline l$.

Let the limit inferior of $\sequence {x_n}$ be $\underline l$.


Then $\left \langle {x_n} \right \rangle$ converges to a limit $l$ if and only if $\overline l = \underline l = l$.


Hence a bounded real sequence converges if and only if all its convergent subsequences have the same limit.


Proof

Sufficient Condition

First, suppose that $\overline l = \underline l = l$.

Let $\epsilon > 0$.

By Terms of Bounded Sequence Within Bounds:

$\exists N_1: \forall n > N_1: x_n < l + \epsilon$

Similarly:

$\exists N_2: \forall n > N_2: x_n > l - \epsilon$

So take $N = \max \set {N_1, N_2}$.

If $n > N$, both the above inequalities hold at the same time.

So $l - \epsilon < x_n < l + \epsilon$ and so by Negative of Absolute Value:

$\size {x_n - l} < \epsilon$

Thus $x_n \to l$ as $n \to \infty$.

$\Box$


Necessary Condition

Now suppose that $\sequence {x_n}$ converges to a limit $l$.

Then by Limit of Subsequence equals Limit of Real Sequence, all subsequences have a limit of $l$ and the result follows.

$\blacksquare$


Sources