Can anyone explain why this series converges? Can anyone explain why this series converges? 
$1+\frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...$
The answer is given, but i do not understand it:
$$**|S_{6n}-S_{3n}|= \frac{1}{3n+1}+\frac{1}{3n+2}-\frac{1}{3n+3}+...+\frac{1}{6n-2}+\frac{1}{6n-1}-\frac{1}{6n}**$$ where $S_{6n}$ and $S_{3n}$ are subseries of $S_n$ partial sums in the respective order, that is
$$**S_{6n}-S_{3n}> \frac{1}{3n+1}+ \frac{1}{3n+4}+...+ \frac{1}{6n-2}>\frac{n}{6n-2}>\frac{1}{6}**$$ 
I highlighted what I dont understand, and would like if someone could simplify, explain..
 A: 
Can anyone explain why this series converges? $~1+\dfrac12-\dfrac13+\dfrac14+\dfrac15-\dfrac16\ldots$

No, because, as others have already said, it diverges. If you want it to converge, you should have $~1+\dfrac12-\dfrac{\color{red}2}3+\dfrac14+\dfrac15-\dfrac{\color{red}2}6\ldots~$ etc. As to why this is so, you can justify it for yourself, by using a reasoning similar to the one presented here.
A: This inequality holds because each of the elements $\frac{1}{3n+1}.....\frac{1}{6n-2}$ is greater or equal to the last term. ie $\frac{1}{3n+1} \gt \frac{1}{6n-2}$ etc for all the terms. So the sum $\frac{1}{3n+1}+\frac{1}{3n+4}...$ (n terms) is definitrly less than $\frac{1}{6n-2}+\frac{1}{6n-2}$...(n terms)..$\frac{1}{6n-2}=\frac{n}{6n-2}$. Is that what you did not understand?
A: No, I can not explain why this series converges. Because it diverges ! Note that the third term is smaller than the second, the sixth is smaller than the fifth, etc. Therefore:
$$S > 1/1 + 1/4 + 1/7 + 1/11 + 1/15....$$
Since the sum of the series on the RHS diverges (logarithmically), so does $S$. 
