Convergence of Limsup and Liminf

Theorem

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

Let the limit superior of $\left \langle {x_n} \right \rangle$ be $\overline l$.

Let the limit inferior of $\left \langle {x_n} \right \rangle$ be $\underline l$.

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

Hence a bounded sequence converges iff all its convergent subsequences have the same limit.

Proof

• 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 $N = \max \left\{{N_1, N_2}\right\}$.

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 $\left|{x_n - l}\right| < \epsilon$.

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

$\blacksquare$

• Now suppose that $\left \langle {x_n} \right \rangle$ converges to a limit $l$.

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

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 5.15 \ (5)$