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To Understanding proof that sequence $x_n =( -1)^{n}$ is not convergent .

Here goes the proof :-

Firstly i assume that sequence converges to x .i.e -1 and 1 lies between $x$-$\epsilon$ and $x+$$\epsilon$ . The length of interval is $2 \epsilon$ which is greater than $2$ (as can be seen by drawing number line ) . so this happens for $\epsilon$ greater than 1 .Because since $2\epsilon$ can only accommodate -1 and 1 if it is greater than 2 i.e if epsilon is greater than 1 .How do i deal with other case when epsilon is less than 1

Thanks

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    $\begingroup$ This reminds me of a joke..."let $\epsilon$ be less than 0". $\endgroup$
    – user223391
    May 1 '15 at 17:10
  • $\begingroup$ It is easy to conclude that this sequence is not a Cauchy sequence. Hence it is not convergent. $\endgroup$ May 1 '15 at 17:12
  • $\begingroup$ Have you learned about Cauchy sequences yet? $\endgroup$
    – Sloan
    May 1 '15 at 17:12
  • $\begingroup$ @Sloan not yet ,no $\endgroup$
    – Taylor Ted
    May 1 '15 at 17:13
  • $\begingroup$ I gave a strong hint. The case when $\epsilon<1$ should be the easier case to prove--and you really don't need cases... just make $\epsilon$ very small. $\endgroup$
    – TravisJ
    May 1 '15 at 17:13
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The definition of a limit is that for any $\epsilon>0$ there exists an $N$ so that if $n\geq N$ then $|x_{n}-x|<\epsilon$. Now, you just need to argue that if $|-1 - x|<\epsilon$ then $|1-x|>\epsilon$ which is a contradiction (to the assumption that $x_n\to x$). Take $\epsilon$ as small as lets you make your argument (smaller than $1$).

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  • $\begingroup$ if mod of -1 -x is less than $\epsilon$ ,then how is mod 1 -x less than $\epsilon$ $\endgroup$
    – Taylor Ted
    May 1 '15 at 17:19
  • $\begingroup$ @K.Dutta, If $|-1-x|<\epsilon$ then $|1-x|>\epsilon$ is the point... we don't want both to be smaller than $\epsilon$ (that's how we get the contradiction). $\endgroup$
    – TravisJ
    May 1 '15 at 18:11
  • $\begingroup$ How come "$|-1 - x|<\epsilon$ then $|1-x|>\epsilon$"? All I can come up with is $|1 + x|<\epsilon$. $\endgroup$
    – abuchay
    Mar 7 '16 at 2:42
  • $\begingroup$ @Tim, it helps to think of what $|-1-x|$ means... it is the distance from $x$ to $-1$. If that distance is small, then $x$ cannot also be close to $1$. And, $|1-x|$ is the distance from $x$ to $1$. Basically, all this is saying is that you cannot be simultaneously close to $1$ and $-1$. $\endgroup$
    – TravisJ
    Mar 7 '16 at 13:51
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You can ignore tha case $\epsilon>1$ right away. Convergence says that for every $\epsilon>0$ we have $|x_n-x|<\epsilon$ for all $n>n_0$ (with $n_0$ dependig on $\epsilon$). Specifically, it should hold for $\epsilon=\frac1{42}$. Since it doesn't, we do not have convergence.

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