Another series involving $\log (3)$ I will show that

$$\sum_{n = 0}^\infty \left (\frac{1}{6n + 1} + \frac{1}{6n + 3} + \frac{1}{6n + 5} - \frac{1}{2n + 1} \right ) = \frac{1}{2} \log (3).$$

My question is can this result be shown more simply then the approach given below? Perhaps using Riemann sums?
Denote the series by $S$ and let $S_n$ be its $n$th partial sum.
\begin{align}
S_n &= \sum_{k = 0}^n \left (\frac{1}{6k + 1} + \frac{1}{6k + 3} + \frac{1}{6k + 5} - \frac{1}{2k + 1} \right )\\
&= \sum_{k = 0}^n \left (\frac{1}{6k + 1} + \frac{1}{6k + 2} + \frac{1}{6k + 3} + \frac{1}{6k + 4} + \frac{1}{6k + 5} + \frac{1}{6k + 6} \right )\\
& \quad - \sum_{k = 0}^n \left (\frac{1}{2k + 1} + \frac{1}{2k + 2} \right ) - \frac{1}{2} \sum_{k = 0}^n \left (\frac{1}{3k + 1} + \frac{1}{3k + 2} + \frac{1}{3k + 3} \right )\\
& \qquad + \frac{1}{2} \sum_{k = 0}^n \frac{1}{k + 1}\\
&= H_{6n + 3} - H_{2n + 2} - \frac{1}{2} H_{3n + 3} + \frac{1}{2} H_{n + 1}.
\end{align}
Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n \frac{1}{k}$. Since $H_n = \log (n) + \gamma + o(1)$ where $\gamma$ is the Euler-Mascheroni constant we see that
$$S_n = \log (6n) - \log (2n) - \frac{1}{2} \log (3n) + \frac{1}{2} \log (n) + o(1) = \frac{1}{2} \log (3) + o(1).$$
Thus
$$S = \lim_{n \to \infty} S_n = \frac{1}{2} \log (3).$$

Simpler, I do not know (except with integration as @J.G. answered).
Otherwise, in order to have have a good approximation of the partial sums, you can start with 
$$s_p=\sum_{n=0}^p \frac 1{a n+b}=\frac{\psi \left(p+1+\frac{b}{a}\right)-\psi
   \left(\frac{b}{a}\right)}{a}$$
and consider that, for large values of $p$,
$$s_p=\frac{\log
   \left({p}\right)-\psi ^{(0)}\left(\frac{b}{a}\right)}{a}+\frac{a+2 b}{2 a^2 p}-\frac{a^2+6 a b+6 b^2}{12a^3 p^2}+\frac{a^2 b+3 a b^2+2 b^3}{6 a^4
   p^3}+O\left(\frac{1}{p^4}\right)$$ which, for your case, would give
$$S_p=\frac 12{\log (3)}-\frac{1}{54 p^2}+\frac{1}{27 p^3}+\cdots$$
