Find $$ 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \dots + \frac{1}{3n + 1} + \frac{1}{3n + 2} - \frac{2}{3n + 3} + \dotsb.$$
I feel to solve this infinite series, you would need to use Taylor Series and probably find the function whose Taylor Series is represented by that, and then obviously calculate. I'm unsure about where to start, what to do to find the function represented by that series. It doesn't really seem to match up with the Taylor Series I'm familiar with. Could I get some help?
Direct method without the use of Taylor series :
For every $N \in \mathbb{N}$, one has \begin{align*} \sum_{n=0}^{N} \dfrac{1}{3n+1} + \dfrac{1}{3n+2} - \dfrac{2}{3n+3} & = \sum_{n=0}^{N} \dfrac{1}{3n+1} + \dfrac{1}{3n+2} + \dfrac{1}{3n+3}- \dfrac{3}{3n+3} \\& = \sum_{n=0}^{N} \left(\dfrac{1}{3n+1} + \dfrac{1}{3n+2} + \dfrac{1}{3n+3}\right)- \sum_{n=0}^{N} \dfrac{1}{n+1} \\ & = \sum_{n=1}^{3N+3} \dfrac{1}{n} - \sum_{n=1}^{N+1} \dfrac{1}{n} \\ & = \ln(3N+3)+\gamma - \ln(N+1)-\gamma + o(1) \\ & = \ln(3)+ o(1) \end{align*}
which leads to $$\boxed{\sum_{n=0}^{+\infty} \dfrac{1}{3n+1} + \dfrac{1}{3n+2} - \dfrac{2}{3n+3} = \ln(3)}$$