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Suppose both $(a_n)$ and $(-a_n)$ have only finitely many peaks of a sequence. Beyond some index $N$, every term has a strictly larger later term and a strictly smaller later term. Starting from $a_N$, construct one strictly increasing and one strictly decreasing monotone subsequence.

If the original sequence had limit of a sequence $L$, both selected subsequences would tend to $L$. The increasing one has a second term strictly above $a_N$, so $L>a_N$; the decreasing one similarly forces $L<a_N$. This contradiction proves nonconvergence. For example, $a_n=(-1{)}^{n}n$ has no peaks of a sequence in either sign and diverges.

For a sequence of real numbers, a common finite limit of a sequence for all monotone subsequences forces the whole sequence to have that limit of a sequence. There is no need to assume a bounded sequence if the hypothesis already includes convergence of every monotone subsequence.

If $a_n$ fails to tend to $L$, select a subsequence with $|a_{n_j}-L|\ge \epsilon$ for some fixed $\epsilon >0$. The monotone subsequence theorem gives a monotone subsequence of this subsequence, which contradicts the common limit of a sequence $L$. Conversely, every subsequence of a convergent sequence has the original limit of a sequence.

For a bounded sequence, the bounded monotone sequence theorem already ensures that all monotone subsequences converge. Only agreement of their limits of sequences remains to be checked.

The Bolzano-Weierstrass theorem states: every bounded sequence of real numbers has a convergent subsequence.

By the monotone subsequence theorem proved in the preceding part, a bounded sequence $(a_n)$ has a monotone subsequence $(b_r)$. It inherits the bounds of $(a_n)$. For completeness, the bounded monotone sequence theorem follows directly from completeness of the real numbers. If $(b_r)$ is increasing, let $L=\sup \{b_r:r\ge 1\}$. For every $\epsilon >0$, the definition of supremum supplies $r_0$ with $L-\epsilon <b_{r_0}\le L$. Thenso $b_r\to L$. If $(b_r)$ is decreasing, apply the increasing case to $(-b_r)$, or use its infimum. Thus in either case

If $a_n\to L$, every subsequence also has limit of a sequence $L$: its increasing indices eventually exceed any prescribed threshold. This applies in particular to every monotone subsequence.

Conversely, suppose $(a_n)$ is a bounded sequence and all its monotone subsequences converge to the same $L$. At least one exists by the monotone subsequence theorem. If $a_n$ did not tend to $L$, there would be ${\epsilon }_0>0$ and a subsequence $(a_{n_j})$ such thatApply the monotone subsequence theorem to $(a_{n_j})$. The resulting further subsequence is also a monotone subsequence of $(a_n)$, so it tends to $L$, contradicting the displayed bound. ThereforeThe bounded monotone sequence theorem ensures that these subsequences have finite limits of sequences; the hypothesis requires those values to agree.

A peak of a sequence of real numbers is an index $k$ whose value is at least every later value. This is a condition on the entire tail, stronger than comparing only adjacent terms. Equality is allowed: every index of a constant sequence is a peak of a sequence.

Infinitely many peaks of a sequence, listed in increasing order of index, give a nonincreasing monotone subsequence. If only finitely many exist, every index beyond the last has a larger later term; repeatedly choosing one gives a strictly increasing monotone subsequence. This proves the monotone subsequence theorem.

A peak of a sequence of $(-a_n)$ supplies a lower bound for all subsequent terms of $(a_n)$. If both $(a_n)$ and $(-a_n)$ have a peak of a sequence, the two selected values bound a common tail. Adding bounds for the finite initial segment proves that $(a_n)$ is a bounded sequence.