Prove that $\lvert\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}\rvert<\frac{1}{n^2}$ Prove that $\lvert\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}\rvert<\frac{1}{n^2}$ and deduce that $1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}$ is convergent. 
Using algebra, the absolute value becomes $\lvert\frac{1}{2n}-\frac{1}{2n-2}\rvert$ which is $\lvert\frac{-(n+1)}{2n(n+1)}\rvert$.
Not entirely sure how to proceed with this proof..
Edit: Incorrect third term, changed to $\frac{1}{2n+2}$.

For the inequality you ask about, use brute force:
$$ \frac{1}{n} - \frac{1}{2n} - \frac{1}{2n-2} = \frac{2(n-1) - (n-1) - n}{2n(n-1)} = \frac{-1}{2n(n-1)}$$
So, your inequality will follow from $2n(n-1) > n^2$, which is equivalent to $n^2 > 2n$, which is true for $n > 2$.
For convergence, you should use the inequality to bound pieces of your series by $\frac{1}{n^2}$ (in absolute value). Once you do this, from convergence of $\sum_n \frac{1}{n^2}$ it follows that the series converges.
