Let $\{x_k\}$ be some sequence of elements from some ordered set (say $\mathbb{R}$). Then for any $n \in \mathbb{N}$, the largest element of the sequence with index less than or equal to $n$ is given by
$$ \max\{ x_k \mid k \le n\} = \max\{ x_1, x_2, \dotsc, x_n\}. $$
The following feels like overkills to me, but if this is an object which is going to be used over and over again, it might be useful to introduces a little extra notation. For example, suppose that $X$ is some ordered set (such as $\mathbb{R}$). The set of all sequences consisting of elements of $X$ is denoted by $X^{\mathbb{N}}$. So we might define
$$ M
: \mathbb{N} \times X^{\mathbb{N}} \to X
: \bigl(n,\{x_k\}\bigr) \mapsto \max\{ x_k \mid k\le n\}.
$$
$M$ is then a function which takes as input a natural number (an index) and some sequence in $X$, and returns the largest element of that sequence with index less than or equal to $n$. In the example given in the question,
$$ M(4, \{ 0, 1, 3, 2, 0, 5, 1, -4, \dotsc \} ) = 3. $$
If the sequence $\{x_k\}$ is fixed, then Alec B-G's suggestion is also reasonable. Thinking of the function $M$ as a function from $\mathbb{N}$ to $X$ (that is, as a sequence of elements of $X$), we could write
$$ M_n = \max\{ x_k \mid k \le n\}. $$
Finally, from a computational standpoint, it might be more efficient to define $M_n = \max\{ M_{n-1}, x_n\}$.