<s>[INST] 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 $\frac{1}{2n+2}$.
 [/INST] $$\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}=\frac{1}{2n}-\frac{1}{2(n+1)}=\frac{1}{2n(n+1)}\leq \frac{1}{n^2}$$
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.
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