public-notes/A Fun Harmonic Series Problem.md

#Math #Calculus

The Problem

If f(x) = \sum _{k \geq 0} a_k x^k, and this series converges for x = x_0, prove:

\sum _{k \geq 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 \geq 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:

\int _0^1 \frac {f(x_0) - f(x_0 y)} {1 - y} dy \\
= \int _0^1 \frac {\sum _{k \geq 0} a_k x_0^k - \sum _{k \geq 0} a_k x_0^k y^k} {1 - y} dy \\
= \sum _{k \geq 1} a_k x_0^k \int _0^1  \frac {1 - y^k} {1 - y} dy

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 \geq 0} a_k x_0^k H_k = \int _0^1 \frac {f(x_0) - f(x_0 y)} {1 - y} dy