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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?

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  • $\begingroup$ Well, show us your work where you got "not Cauchy", then we can check for it for you. $\endgroup$ – 0XLR Apr 8 at 2:51
  • $\begingroup$ @ZeroXLR just edited my question, check please. Thanks $\endgroup$ – analysis1 Apr 8 at 2:55
  • $\begingroup$ 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 $\endgroup$ – 0XLR Apr 8 at 2:58
  • $\begingroup$ It converges, so it is Cauchy. $\endgroup$ – dmtri Apr 8 at 3:04
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    $\begingroup$ 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. $\endgroup$ – fleablood Apr 8 at 6:23
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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?

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

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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 $|x_n - x_m| \le |x_n| + |x_m| < \frac 12 \epsilon + \frac 12 \epsilon = \epsilon$.

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