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 evaluating sums of the form $\sum\limits_{i=1}^n \frac1{ai+b}$?
 A: 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.
A: $$\begin{align}
\frac11 + \frac13 + \frac15 + \cdots + \frac1{2n-1} &= \left( \frac11 + \frac12 + \frac13 + \frac14 + \cdots + \frac1{2n} \right) - \left( \frac12 + \frac14 + \cdots + \frac1{2n} \right)\\ & = H_{2n} - \frac12H_n
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
