A sequence $(s_n)$ is monotonically nonincreasing if

$$ s_m \le s_n \text{ for all } m > n $$

Similarly, a real function $f(x)$ is monotonically nonincreasing if

$$ f(x) \le f(y) \text{ for all } x > y $$

Compare this to monotonically decreasing.

Conflict note. In some contexts, such as [1], this is called monotonically decreasing (in turn, our ``monotonically decreasing'' is called ``strictly decreasing''). This is unfortunately counter-intuitive, since a sequence or function that is ``flat'' (such as $f(x) = 1$ is somehow ``decreasing.'' Beware!

Examples

• $(s_n) = 1, 0, -1, -2, \ldots$ is monotonically nonincreasing. It is also monotonically decreasing.
• $(s_n) = 1, 1, 1, 1, \ldots$ is nonincreasing but not monotonically decreasing.
• $(s_n) = (\frac{1}{n+1})$ is nonincreasing (note that $n$ is nonnegative).
• $(s_n) = 1, 1, 2, 1, 1, \ldots$ is not nonincreasing. It also happens to fail to be monotonically nondecreasing.
• $(s_n) = 1, 2, 3, 4, 5, \ldots$ is not nonincreasing, rather it is nondecreasing (and monotonically increasing).

Bibliography

1

``monotonically decreasing,'' from the NIST Dictionary of Algorithms and Data Structures, Paul E. Black, ed.