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 $\frac{1}{2n+2}$.

I think you have it wrong there. The absolute value becomes $\dfrac{1}{2n^2-2n}$ which is $\dfrac{1}{2n^2-2n}=\dfrac{1}{n^2+n^2-2n}<\dfrac{1}{n^2}$ when $n^2>2n$.
