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.