A Fun Harmonic Series Problem

Math Calculus

The Problem

If $f(x)={\sum }_{k\ge 0}{a}_k{x}^k$, and this series converges for $x=x_0$, prove:

$$\sum _{k\ge 0}{a}_k{x}_0^k{H}_k={\int }_0^1\frac{f(x_0)-f(x_{0}y)}{1-y}dy$$

where $H_k$ is defined to be the partial sums of the harmonic series ($H_0=0$, $H_k={\sum }_{i=1}^k\frac{1}{i}$ for $k\ge 1$).

(from The Art of Computer Programming)

Solution

Although this problem might seem intimidating with a power series involving the harmonic numbers on the LHS and a summation function inside an integral on the RHS, it is fairly trivial to bring out the summation and express the RHS as a power series:

$$\begin{gathered} {\int }_0^1\frac{f(x_0)-f(x_{0}y)}{1-y}dy \\ ={\int }_0^1\frac{{\sum }_{k\ge 0}{a}_k{x}_0^k-{\sum }_{k\ge 0}{a}_k{x}_0^k{y}^k}{1-y}dy \\ =\sum _{k\ge 1}{a}_k{x}_0^k{\int }_0^1\frac{1-y^k}{1-y}dy \end{gathered}$$

The integral factor on the last step is now merely Euler’s integral representation for the harmonic numbers, which is easily proven by the simple fact that $\frac{1-y^k}{1-y}={\sum }_{i=0}^{k-1}{y}^i$. Therefore:

$$\sum _{k\ge 0}{a}_k{x}_0^k{H}_k={\int }_0^1\frac{f(x_0)-f(x_{0}y)}{1-y}dy$$