<s>[INST] $1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$ 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?
 [/INST] $$\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}=\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{4n}-\frac{1}{4n}=\frac{1}{4n-3}-\frac{1}{4n-2}+\frac{1}{4n-1}-\frac{1}{4n}+\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$ 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
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