# Finite sum of reciprocal odd integers

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 $\sum\limits_{i=1}^n \frac1{ai+b}$?

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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

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

$$H_n=\gamma+\psi(n+1)$$

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

$$H_{n-\frac12}=\gamma+\psi\left(n+\frac12\right)$$

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

$$\psi(2z)=\log\,2+\frac12\left(\psi(z)+\psi\left(z+\frac12\right)\right)$$

which, when expressed in harmonic number terms, is

$$H_{2n-1}=\log\,2+\frac12\left(H_{n-1}+H_{n-\frac12}\right)$$

Thus,

\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.

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\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}

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