$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$ conditionally convergent $$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$$
I want to show first that $S_{3n}$, $S_{3n+1}$ and $S_{3n+2}$ converges to the same limit, I show 
$$S_{3n} = (1 + {1 \over 3} - {1 \over 2}) + ({1 \over 5} + {1 \over 7} - {1 \over 4}) + ({1 \over 9} + {1 \over 11} - {1 \over 6}) + ... + ({1 \over 4n-3} + {1 \over 4n-1} - {1 \over 2n})$$
but how to proceed from here?

The series is 
$$\sum_{n=1}^\infty\left(\frac1{4n-3}+\frac1{4n-1}-\frac1{2n}\right)=\sum_{n=1}^\infty\frac{8n-3}{2n(4n-1)(4n-3)}$$
and this thing is a converging positive series, since
$$\frac{8n-3}{2n(4n-1)(4n-3)}\le\frac4{(4n-1)(4n-3)}$$
