Finite sum of reciprocal odd integers

Question

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 evaluating sums of the form $\sum\limits_{i=1}^n \frac1{ai+b}$?

Answer 1

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

Answer 2

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.

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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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{i = 1}^{n}{1 \over 2i - 1} & = \sum_{i = 0}^{n - 1}{1 \over 2i + 1} = {1 \over 2}\sum_{i = 0}^{\infty} \pars{{1 \over i + 1/2} - {1 \over i + 1/2 + n}} \\[5mm] &= {1 \over 2}\bracks{\Psi\pars{n + {1 \over 2}} - \Psi\pars{1 \over 2}} \\[5mm] & = {1 \over 2}\ \underbrace{\braces{\Psi\pars{\bracks{\color{red}{n - {1 \over 2}}} + 1} + \gamma}}_{\ds{H_{n - 1/2}}}\ -\ {1 \over 2}\ \underbrace{\bracks{\Psi\pars{\color{red}{1 \over 2}} + \gamma}} _{\ds{\int_{0}^{1}{1 - t^{\color{red}{1/2} - 1} \over 1 - t}\,\dd t}} \\[5mm] & = {1 \over 2}\,H_{n - 1/2} - {1 \over 2}\int_{0}^{1}{1 - t^{- 1} \over 1 - t^{2}}\, 2t\,\dd t = {1 \over 2}\,H_{n - 1/2} + \int_{0}^{1}{\dd t \over 1 + t} \\[5mm] & = \bbox[10px,#ffd,border:1px groove navy]{{1 \over 2}\,H_{n - 1/2} + \ln\pars{2}} \end{align}