$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$ 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?

$$|S_{3n+1}-S_{3n}| \leq \frac{1}{n}$$
$$|S_{3n+2}-S_{3n}| \leq \frac{1}{n}$$
So if $(S_{3n})$ converges, all the three converges
No to prove that $(S_{3n})$ converges, all you need to do is to write $\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}$ over the same denominator and use a comparaison.
