Sequence And Series,is It?

Note first that $\dfrac{1}{3n + a} = \displaystyle\int_{0}^{1} x^{3n + a - 1} dx$ for integers $n \ge 0, a \ge 1.$

Looking then at a partial sum, we have that

$\displaystyle\sum_{n=0}^{m} (\dfrac{1}{3n + 1} - \dfrac{1}{3n + 2}) = \sum_{n=0}^{m} \int_{0}^{1} (x^{3n} - x^{3n + 1}) dx = \int_{0}^{1} \sum_{n=0}^{m} (x^{3n} - x^{3n + 1}) dx,$

where the exchange of summation and integral signs is valid since we have convergence for $|x| \le 1.$ Now for $|x| \lt 1$ we have that

$\displaystyle\sum_{n=0}^{m}(x^{3n} - x^{3n + 1}) = \sum_{n=0}^{m} x^{3n}(1 - x) = \dfrac{1 - x^{3m+1}}{1 - x^{3}}*(1 - x) = \dfrac{1 - x^{3m + 1}}{1 + x + x^{2}},$

so our integral now becomes $\displaystyle\int_{0}^{1} \dfrac{1 - x^{3m + 1}}{1 + x + x^{2}} dx$ ,

which goes to $\displaystyle\int_{0}^{1} \dfrac{1}{1 + x + x^{2}} dx$ as $m \rightarrow \infty.$

(I'm getting a bit fast and loose with endpoint behavior as $x \rightarrow 1$ and with eliminating the dominated $x^{3m + 1}$ term from the integral so casually, but this solution is getting long enough as it is, so.....)

This integral can then be solved by trig substitution. Since $x^{2} + x + 1 = (x + \frac{1}{2})^{2} - \frac{1}{4} + 1 = (x + \frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}$ , we let

$x + \frac{1}{2} = \frac{\sqrt{3}}{2}\tan(\theta) \Longrightarrow dx = \frac{\sqrt{3}}{2}\sec^{2}(\theta)$ ,

transforming the (indefinite) integral to

$\displaystyle\int \dfrac{\frac{\sqrt{3}}{2}\sec^{2}(\theta)}{\frac{3}{4}\sec^{2}(\theta)} d\theta = \dfrac{2}{\sqrt{3}}\theta = \dfrac{2}{\sqrt{3}}\tan^{-1}(\frac{2x + 1}{\sqrt{3}}),$

which, when evaluated from $x = 0$ to $x = 1$ gives us

$\dfrac{2}{\sqrt{3}}(\tan^{-1}(\sqrt{3}) - \tan^{-1}(\dfrac{1}{\sqrt{3}})) = \dfrac{2}{\sqrt{3}}(\dfrac{\pi}{3} - \dfrac{\pi}{6}) = \dfrac{\pi}{3\sqrt{3}} = 0.604599788....$

So $\lfloor 10000*\gamma \rfloor = \boxed{6045}$ .