# Is the following sequence $x_{n}=\frac{1 +(-1)^n}{n}$ Cauchy?

Is the following sequence $$x_{n}=\frac{1 +(-1)^n}{n}$$ Cauchy? I got not Cauchy, but would appreciate someone to check this. Thank you! So I got the sequence is bounded by 2 and that $$abs(x_n-x_{n+1})<=2$$ then if we pick $$\epsilon=1/2$$ then the sequence is not Cauchy... is my proof sufficient?

• Well, show us your work where you got "not Cauchy", then we can check for it for you. – 0XLR Apr 8 at 2:51
• @ZeroXLR just edited my question, check please. Thanks – analysis1 Apr 8 at 2:55
• Ok, how does being bounded above by $2$ prevent the sequence elements from not get any closer than $\epsilon = 1 / 2$? To conclude that, you will need a lower bound not an upper bound – 0XLR Apr 8 at 2:58
• It converges, so it is Cauchy. – dmtri Apr 8 at 3:04
• 1) If $\epsilon = \frac 12$ and $n,m> 4$ then $|x_n-x_m| < \frac 12 = \epsilon$ so nothing you said makes any sense and 2) you can't show something for one epsilon; you must show it for all. – fleablood Apr 8 at 6:23

Hint: Recall that a convergent sequence is automatically Cauchy. Next, note that the triangle inequality yields $$|x_n| \leq \frac{1 + |(-1)^n|}{n} =\frac{2}{n}.$$ From here can you show that $$\lim{x_n}$$ exists?

$$|x_n-x_m| \leq \frac 1 n +\frac 1 n +\frac 1 m +\frac 1 m <\epsilon$$ whenever $$n, m \geq \frac 4 {\epsilon}$$.

If $$n$$ is even then $$x_n = \frac {1+1}n = \frac 2n$$ and if $$n$$ is odd then $$x_n = \frac {1-1}n = 0$$.

It should be well known that $$\frac 1n \to 0$$ so $$\frac 2n\to 0$$ and, of course $$0\to 0$$ so $$x_n \to 0$$ and converging sequences are Cauchy so this should be Cauchy.

Just as we prove that $$b_n = \frac 1n$$ is a cauchy sequence by saying for any $$\epsilon > 0$$ then if we let $$N = \frac 2{\epsilon}$$ then if $$n,m > N$$ then $$\frac 1n < \frac 1N = \frac \epsilon 2$$ and $$\frac 1m <\frac \epsilon 2$$ so $$|b_n - b_m|=|\frac 1n - \frac 1m| \le |\frac 1n| + |\frac 1n| < \frac \epsilon 2 + \frac \epsilon 2$$...

We prove $$x_n$$ is cauchy by pointing out for any $$\epsilon > 0$$ if we let $$N = \frac 4\epsilon$$ then if $$n,m > N$$ then if $$n$$ is even then $$x_n =\frac 2{n}< \frac 2N = \frac 12\epsilon$$ and if $$n$$ is odd then $$x_n = 0 < \frac 12 \epsilon$$ so either way $$x_n < \frac 12 \epsilon$$.

Same thing for $$x_m$$ and $$|x_n - x_m| \le |x_n| + |x_m| < \frac 12 \epsilon + \frac 12 \epsilon = \epsilon$$.