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 $\frac{1}{2n+2}$.
 A: $$\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}=\frac{1}{2n}-\frac{1}{2(n+1)}=\frac{1}{2n(n+1)}\leq \frac{1}{n^2}$$
Now we have
$$0\leq1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\cdots=\sum_{n=1,n \,\mathrm{odd}}^\infty\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}\leq\sum_{n=1}^\infty\frac{1}{n^2}$$
the series is positive and bounded above so it's convergent.
A: $$
\begin{array}{rcl}
\left| \frac{1}{n}-\frac{1}{2n}-\frac{1}{2n-2} \right| & < & \frac{1}{n^2} \\
\left| \frac{1}{2n}-\frac{1}{2n-2} \right|& < & \frac{1}{n^2} \\
\left| \frac{-2}{(2n)(2n-2)} \right|& < & \frac{1}{n^2} \\
\frac{1}{2n^2-2n}& < & \frac{1}{2n^2-n^2} \\
\end{array}
$$
True for all integer $n>1$.
A: 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.
A: 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 $n^2>2n$.
