Let an = {7 + 4/n if n is even, 8 - 1/n if n is odd}. Prove that the sequence {an} does not converge by showing that it's not a Cauchy sequence.

This is what I have so far.

Let $\epsilon$ > 0. For this sequence to be Cauchy then there exists a natural number N such that,

|7 + 4/n| < $\epsilon$/2 and |8 - 1/m| < $\epsilon/2$ $\forall$n,m ≥N $\iff$ |7 + 4/n - 8 + 1/m|<$\epsilon$

From here I don't know what to do. I believe I'm supposed to choose a specific $\epsilon$ and show that there does not exist an N that satisfies this. I thought I would choose $\epsilon$ = 1, then |7 + 4/n| < 1/2 and |8 + 1/m| < 1/2 but I'm not sure where to go from here or if I'm even on the right track.

Also, I know that the 2 sequences themselves don't converge, and if they're subsequences of {an} then {an} wouldn't converge but how do I show this in the proof?

Thanks for any help, sorry for my formatting, couldn't quite figure that out either.

  • $\begingroup$ Is it $a_n=7+4/n$ if $n$ is even and $8-1/n$ if $n$ is odd? $\endgroup$ – Zero Jan 22 '15 at 2:45
  • $\begingroup$ yes, that is correct. $\endgroup$ – Mary Martinez Jan 22 '15 at 3:00

Take $\epsilon=1/2$ for example, suppose there is an $N$ such that $\forall n\forall m:n,m\ge N\implies |a_n-a_m|\le 1/2$. Pick up $n$ odd and so big that $n\ge 20$ and $n\ge N$. We then have $a_n-a_{n+1}=1-1/n-4/(1+n)>1/2$ and we found our desired contradiction.

  • $\begingroup$ Could you explain why $\an − an+1 = 8 − 1/n − 4/(1+n)$? If an is even and n+1 is odd (assuming I've chosen n=20), then wouldn't it be $\an - an+1 = 7 + 4/n - 8 + 1/n$? I'm just trying to understand how you arrived at that. Also not sure how you got $\4/(1+n)$ $\endgroup$ – Mary Martinez Jan 22 '15 at 3:17
  • $\begingroup$ Sorry I made a mistake, it is now fine, $a_n-a_{n+1}=8-1/n-(7+4/(n+1))=1-1/n-4/(1+n)$. $\endgroup$ – Zero Jan 22 '15 at 12:19
  • $\begingroup$ Thank you so much, now it makes sense!! $\endgroup$ – Mary Martinez Jan 22 '15 at 16:22

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