Is there any way to differentiate such function? Let $S$ be a set.
If I had a bijection $f$ mapping each element $n\in \mathbb{N}$ to an element $s \in S$ such that:
$$s = f(n) = \sum^{n}_{k=1} {1\over k}$$
Is the function differentiable in respect to $n$?
I know it's an odd question, but it's stuck in my head.
 A: No. Differentiability is only defined within a certain subset of functions, specifically ones where you can reasonably compute limits. What "reasonably" means depends a bit on the context, but usually requires that it be defined on a field (generally a real field), which $\mathbb{N}$ is not. There are ways to extend the notion of a limit to such places, with ultra filters, but such discussion is not contained within what is normally meant by "derivative" and "limit", so if you're interested in those kinds of behaviors I would recommend you ask specifically about if ultra filters can be used to define a notion of a derivative on discrete sets.
A: Calculus is normally done with functions from $\mathbb{R} \rightarrow \mathbb{R}$. So no, you can't use the derivative you're familiar with on $f$. I can tell you that just by knowing the domain and the codomain (range), without even looking at how your function is defined.
You can see why this doesn't work by looking at the definition of the derivative:
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$
The limit tells us that $h$ is going to get very small. But if your function $f$ is only defined on natural numbers, how do you compute $f(x + .001)$?
Here's something we can do for functions from $\mathbb{N} \rightarrow \text{something}$. I feel it preserves many of the main ideas of the derivative. Define $\Delta f(x)$ by:
$$\Delta f(x) = \frac{f(x + 1) - f(x)}{1} = f(x + 1) - f(x)$$
This is just the previous formula with $h = 1$ and without the limit. It's known as the "discrete derivative" or "forward difference" (backward difference would be $f(x) - f(x-1)$). Using it, we can take derivative of sequences: for example, if $f(n) = n^2$, then
$$\Delta f(n) = (n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$$
That's showing that the distance between any two square numbers is always odd.
So we can take the discrete derivative of your function:
$$\Delta f(n) = f(n+1) - f(n) = \sum_{k=1}^{n+1} \frac{1}{k} - \sum_{k=1}^n \frac{1}{k} = \frac{1}{n+1}$$
