Peak of a sequence ($a_k\ge a_m\text{for every }m>k$)

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.

Suppose $k_1<k_2<\cdots$ are peaks of a sequence $(a_n)$ and $m_1<m_2<\cdots$ are peaks of a sequence $(-a_n)$. Then $(a_n)$ converges precisely when these two subsequences have a common finite limit of a sequence $L$.

Necessity follows because every subsequence inherits the limit of a sequence of a convergent sequence. For sufficiency, choose one selected value from each subsequence within $\epsilon$ of $L$. Every term after both selected indices lies between the two selected values:This is exactly $a_n\to L$.

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.