How to prove $\frac{(-1)^{(n-1)}}n-1$ is Cauchy

Question

So I've got to prove $\frac{(-1)^{(n-1)}}n-1$ is a Cauchy sequence, but I can't do that if I can't simplify it to the point at which 1 is the numerator (so I can cross multiply with $\frac{2}{\epsilon}$), which I'm not sure you can.

I get as far as $|\frac{n+(-1)^n}{n}|+|\frac{m+(-1)^m}{m}| < \epsilon$ but have no idea how to simplify further.

Prove that it converges.Every convergent sequence is cauchy. — Grada Gukovic, Dec 09 '19 at 00:42

score 1 · Answer 1 · answered Dec 09 '19 at 00:43

1

$$\left|\frac{(-1)^{n-1}}{n}-1 - \frac{(-1)^{m-1}}{m}+1\right| \leq 2 \max\left\{\frac{1}{n}, \frac{1}{m}\right\},$$ hence if $n,m > \frac{2}{\epsilon}$, you have $$\left|\frac{(-1)^{n-1}}{n} - \frac{(-1)^{m-1}}{m}\right| < \epsilon.$$

answered Dec 09 '19 at 00:43

Dzoooks

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rims · Accepted Answer · 2019-12-13T08:48:10.050

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Let $a_n = \frac{(-1)^{n-1}}{n} - 1$

Pick some $m, n \in \mathbb{N}$ so that $$|a_n - a_m|$$ $$ = \Big |\frac{(-1)^{n-1}}{n} - 1 - \frac{(-1)^{m-1}}{m} + 1\Big|$$ $$ = \Big|\frac{(-1)^{n-1}}{n} - \frac{(-1)^{m-1}}{m}\Big| $$ $$\leq \Big|\frac{(-1)^{n-1}}{n} \Big | + \Big |\frac{(-1)^{m-1}}{m} \Big|$$ $$ = \Big| \frac{1}{n} \Big| + \Big| \frac{1}{m} \Big|$$

Given any $\epsilon > 0$, choose $N = 2 * \lceil \frac{1}{\epsilon} \rceil$ so that $n, m > N$ $\implies $

$$|a_n - a_m| \leq \Big| \frac{1}{n} \Big| + \Big| \frac{1}{m} \Big| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

edited Dec 13 '19 at 08:48

answered Dec 09 '19 at 00:48

rims

2,657

So you can just simplify $(-1)^{n-1}$ to 1? – James M Dec 09 '19 at 00:51
$(-1)^{n-1}$ is either $-1$ or $+1$ so $|(-1)^{n-1}|$ is always $1$. – rims Dec 09 '19 at 00:55
Also, $\Big | \frac{a}{b} \Big | = \frac{|a|}{|b|} = \Big | \frac{a}{|b|} \Big | = \Big | \frac{|a|}{b} \Big | $ if that's what is confusing you. – rims Dec 09 '19 at 00:57
Thanks! All good now. – James M Dec 09 '19 at 00:59

How to prove $\frac{(-1)^{(n-1)}}n-1$ is Cauchy

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