Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Mathematica tells me that $\sum\limits_{i=1}^n \frac1{2i-1}$ is equal to $\frac12 H_{n-1/2}+\log\,2$, where $H_n$ is a harmonic number.

Why is this true? Is there a general strategy for solving sums of the form $\sum\limits_{i=1}^n \frac1{ai+b}$?

share|cite|improve this question
rational=rational+irrational, eh? – Alex Youcis May 5 '12 at 3:19
@AlexYoucis Fractional Harmonics evaluate in terms of Digamma. – Pedro Tamaroff May 5 '12 at 3:24
up vote 4 down vote accepted

To elaborate on Peter's comment: the harmonic numbers $H_n=\sum\limits_{k=1}^n\frac1{k}$ and the digamma function $\psi(z)=\frac{\Gamma^\prime(z)}{\Gamma(z)}$ satisfy the relationship


($\gamma$ is the Euler-Mascheroni constant), which means that


Now, there is the duplication theorem (which can be derived from the duplication theorem for the gamma function):


which, when expressed in harmonic number terms, is



$$\begin{align*} \sum_{k=1}^n \frac1{2k-1}&=\log\,2+\frac12 H_{n-\frac12}\\ &=\log\,2+\frac12(2H_{2n-1}-H_{n-1}-2\log\,2)\\ &=\frac12(2H_{2n-1}-H_{n-1})\\ &=\frac12(2H_{2n}-\frac1{n}-\left(H_{n}-\frac1{n}\right))=\frac12(2H_{2n}-H_n) \end{align*}$$

which is what Marvis got through simpler means.

In general, through formal manipulation:

$$\begin{align*} \sum_{k=1}^n \frac1{ak+b}&=\frac1{a}\sum_{k=1}^n \frac1{k+\frac{b}{a}}\\ &=\frac1{a}\sum_{k-\frac{b}{a}=1}^n \frac1{k}=\frac1{a}\sum_{k=\frac{b}{a}+1}^{n+\frac{b}{a}} \frac1{k}\\ &=\frac1{a}\left(\sum_{k=1}^{n+\frac{b}{a}} \frac1{k}-\sum_{k=1}^{\frac{b}{a}} \frac1{k}\right)\\ &=\frac1{a}\left(H_{n+\frac{b}{a}}-H_{\frac{b}{a}}\right) \end{align*}$$

and one might be able to use the multiplication theorem to express fractional values of harmonic numbers as linear combinations of harmonic numbers of integer argument.

share|cite|improve this answer

$$\begin{align} \frac11 + \frac13 + \frac15 + \cdots + \frac1{2n-1} &= \left( \frac11 + \frac12 + \frac13 + \frac14 + \cdots + \frac1{2n} \right) - \left( \frac12 + \frac14 + \cdots + \frac1{2n} \right)\\ & = H_{2n} - \frac12H_n \end{align} $$

share|cite|improve this answer
Thanks -- although I accepted the other answer for its completeness, I like the simplicity of this argument. – tba May 6 '12 at 0:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.