# Prove $\{(-2)^n\}$ is not Cauchy by using the definition

I've gotten that $|x_m - x_n|$ = 0 if $n=m$. I can't find the pattern to explain when $n+m$ is odd.

• let $\epsilon =1$ and try to apply the definition of Cauchy sequence. In particular, note that $|x_m-x_{m+1}| >\epsilon$ for any $m$. – Surb Dec 9 '15 at 19:56
• "I've gotten that $|x_m - x_n|$ = 0 if n+m is even." That's false. Try some values. – quid Dec 9 '15 at 19:59
• ^ Oops, check edit – p3ngu1n Dec 9 '15 at 20:00
• Note that if $m>n$, then $$|(-2)^m-(-2)^n|\geq 2^m-2^n =2^n(2^{m-n}-1)\geq 2^n$$ – MPW Dec 9 '15 at 20:01

It's enough to find $n, n+1$ so that $x_n - x_{n+1}$ is not arbitrarily small even if picking $n$ large enough.
But $|(-2)^n - (-2)^{n+1}| = |2^n (1 - (-2))| = 3 \cdot 2^n$, so that for $\varepsilon < 3$, you cannot make the difference small enough by changing $n$ (it will always be $\geq 3$).