# Summation of Power Series by Harmonic Sequence

## Theorem

Consider the power series:

$\map f x = \displaystyle \sum_{k \mathop \ge 0} a_k x^k$

Let $\map f x$ converge for $x = x_0$.

Then:

$\displaystyle \sum_{k \mathop \ge 0} a_k {x_0}^k H_k = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y$

where $H_n$ denotes the $n$th harmonic number.

## Proof

 $\displaystyle \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y$ $=$ $\displaystyle \int_0^1 \dfrac {\sum_{k \mathop \ge 0} a_k {x_0}^k - \sum_{k \mathop \ge 0} a_k {x_0}^k y^k} {1 - y} \rd y$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop \ge 0} a_k {x_0}^k \int_0^1 \dfrac {1 - y^k} {1 - y} \rd y$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop \ge 0} a_k {x_0}^k \int_0^1 \paren {1 + y + y^2 + \cdots + y^{k - 1} } \rd y$ Sum of Geometric Sequence $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop \ge 0} a_k {x_0}^k \intlimits {y + \dfrac {y^2} 2 + \dfrac {y^3} 3 + \cdots + \dfrac {y^k} k} 0 1$ Primitive of Power $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop \ge 0} a_k {x_0}^k \paren {\paren {1 + \dfrac {1^2} 2 + \dfrac {1^3} 3 + \cdots + \dfrac {1^k} k} - \paren {0 + \dfrac {0^2} 2 + \dfrac {0^3} 3 + \cdots + \dfrac {0^k} k} }$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop \ge 0} a_k {x_0}^k H_k$ Definition of Harmonic Numbers

$\blacksquare$