Solving an equation with a function

Question

So the problem is to find $f(x)$ such that: $$f(x+1)-f(x)=1/(x+1)$$

I have found that $\ln x$ is a good approximation for large values of $x$.

$f(x)$ not differentiable at $x=-1$.

Answer 1

This is a simple recurrence, that can be written

$$f(x)=f(x-1)+\frac1x.$$

By induction,

$$f(x)=f(x-n)+\sum_{k=0}^{n-1}\frac1{x-k}.$$

If we assume the initial condition that $f$ is known in $[0,1)$, we have

$$f(x)=f(\{x\})+\sum_{k=0}^{\left\lfloor x\right\rfloor-1}\frac1{x-k}.$$

For large $x$, it indeeds tends to an Harmonic series, with a perturbation term.