# Finite derivative of the harmonic series

In Knuth's Concrete Mathematics he represents the famous quicksort algorithm in computer science as a infinite sum then shows that sum can be simplified to being essentially harmonic. I want to explore this sum more by taking a finite derivative of the function but I have not taken any discrete math courses and am struggling with notation.

My question:

Suppose, $$f(n)=\sum_{n=0}^\infty \frac{1}{n+1}$$

Now suppose we take the generalized finite difference of that function, $$\Delta _h^\mu [f](x)=\sum_{x=0}^\infty \mu_k f(x+kh)$$

Have I set this up properly?

I want to explore the finite difference of this series. If I have not set this up correctly, might you point me in the right direction of what I might read to learn how to do this myself. Thank you.

• $f$ is not a function of $n$, or is it? – Math-fun Feb 22 '16 at 11:59
• It is ... $n$ appears both as a free and a bounded variable ... not a very good idea. Besides this: As the harmonic series does not converge ... $f \equiv \infty$. – martini Feb 22 '16 at 12:00
• To my understanding $f$ is a function of n, otherwise this problem would not work. But I may be confused. – jake mckenzie Feb 22 '16 at 12:00
• Part of my reason of asking this martini is that I wanted to explore this: I know the harmonic series diverge, but if I differentiate it to infinity, does that series converge? – jake mckenzie Feb 22 '16 at 12:04
• I made some edits. I don't know if that fixes the problem. – jake mckenzie Feb 22 '16 at 12:06

You probably mean $$f(n)=\sum_{k=1}^n \frac{1}{k}\approx \ln n+\gamma$$ As a sum is basically a discrete integral, the finite derivative of a sum over the upper bound gives you back the summand $f'(n)\asymp \frac{1}{n}$. More formal treatment of differentiation of discrete sums is found here.
Anyway, I think you can deal with the finite difference implementation now: the difference between two partial sums for different $n$ shouldn't be a problem.