Summation of Odd Reciprocals in terms of Harmonic Numbers

Theorem

$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = H_{2 n} - \dfrac {H_n} 2$

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

$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}$

$=$

$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} + \sum_{k \mathop = 1}^n \dfrac 1 {2 k} - \sum_{k \mathop = 1}^n \dfrac 1 {2 k}$

$\ds $

$=$

$\ds \sum_{k \mathop = 1}^{2 n} \dfrac 1 k - \dfrac 1 2 \sum_{k \mathop = 1}^n \dfrac 1 k$

$\ds $

$=$

$\ds H_{2 n} - \dfrac {H_n} 2$

$\blacksquare$

Donald E. Knuth originally published the following result in The Art of Computer Programming: Volume 1: Fundamental Algorithms, 3rd ed. ($1997$) as a solution to the exercise to evaluate $\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}$:

$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = H_{2 n - 1} - \dfrac {H_{n - 1} } 2$

The online errata sheet, found via https://www-cs-faculty.stanford.edu/~knuth/taocp.html, offers the simpler result as a replacement.