Rate of change converging for a discrete function?

Let's say I have a function, say $f(x)$. However, $x$ is not continuous, it can only be integers. So, I only have definitions for $f(1),f(2),f(3)$, etc. If you were to plot out f of all the integers, and connect the dots, it would look like a monotone decreasing function. Can I call this function monotone even though it's not continuous? And is there a formal way of saying "the continuous, monotone function that would be created by connecting the dots of $f(x)$?

Basically, I want to show that $f(x)$ decreases at a decreasing rate. I can show that $f(x+1) < f(x)$ and that $f(\infty) =$ a constant. Can I do this with an integer-based function?

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JandR, is it that you do not care about the function at any non-integer $x$? Is the function defined for all reals or is it defined at only the integers? – Srivatsan Sep 18 '11 at 4:10
@Srivatsan: Hmm.. I mean that in my case, $f$ is really a recursive function, and $x$ is the number of times of recursion. So I have some inner function say $g(y)$ and $f(1) = g(y)$, $f(2) = g(g(y))$, $f(3) = g(g(g(y)))$ etc. I say $f$ is only defined for integer values of $x$ because I don't know how to interpret a partial recursion. Does that make sense? But, I still want to be able to treat $f(x)$ as continuous, or at least make inferences about its rate-of-rate of change and convergence.. – OctaviaQ Sep 18 '11 at 4:14
JandR Functions $f : \mathbb N \to \mathbb R$ are also called sequences since you can identify it with the sequence $f(1), f(2), \ldots$. For such functions, monotonicity, convergence, limits etc. are all well-defined. – Srivatsan Sep 18 '11 at 4:20
@Srivatsan: Oh, ok, thanks. I will go read up on sequences. I hadn't thought of that! – OctaviaQ Sep 18 '11 at 4:24

As for talking about extending the function, I would call it the piecewise affine extension of $f$ to the real numbers which is affine on the interval $[n,n+1]$ for each integer $n$.
Finally, if you want to talk about this function (the piecewise affine extension of $f$) decreasing more slowly, you are talking about the concavity of the function. Specifically, you are saying that the function is decreasing and convex. Convexity does not follow even eventually from the fact that it is decreasing and bounded below. Take the example that $f(n)=\frac{1}{\lfloor (n+1)/2 \rfloor}$ for $n>0$ and $f(n)=1$ otherwise.