$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$ 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?
 A: 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
$$\frac{8n-3}{2n(4n-1)(4n-3)}\le\frac4{(4n-1)(4n-3)}$$
A: $$\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}=\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{4n}-\frac{1}{4n}=\frac{1}{4n-3}-\frac{1}{4n-2}+\frac{1}{4n-1}-\frac{1}{4n}+\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$ so $$\lim _{n\to \infty}S=\frac{3}{2}\lim _{n\to \infty}\sum _n\left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$ and $$\log (x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}..$$ so $$\log(2)=\log(1+1)=\sum \left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$ so $S=\frac{3}{2}\log (2)$ Hope it will help
A: $$|S_{3n+1}-S_{3n}| \leq \frac{1}{n}$$
$$|S_{3n+2}-S_{3n}| \leq \frac{1}{n}$$
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.
