$f(x)=f(x-1)+\dfrac{1}{x}$

105 Views Asked by At

I have the $x$th partial sum of the harmonic series, $\sum_{k=1}^{x}{\dfrac{1}{k}}$, and I want to find a function that is equivalent for integer inputs greater than $0$ so I can take it's derivative which is another question. So what I did is I wrote down the functional equation:

$$f(x)=f(x-1)+\dfrac{1}{x}\;\;\forall x\in \mathbb{N}\setminus\{0,1\}\land f(1)=1$$

I don't even know if such a function exists. I think it might have to do something with an integral of $\dfrac{-1}{x^2}$ or maybe $x$ is a bound of a definite integral. I tried $\int_{1-x}^{x}\dfrac{-1}{x^2}dx$, but it only adds $\dfrac{1}{x}$ to $\dfrac{1}{x-1}$. I know I can sum the integral over some $k$, but then I am back again where I started.

Any help would be appreciated. Thank you.