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.
Best Answer
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.