Verification of proof of nonconvergence

Question

I'm not sure if my proof is sound or not, I was wondering if anyone could verify.

Prove that the sequence $\{a_n\}_n $ given by $a_n = \frac{(-1)^n\;n+1}{3n}$ does not converge.

Proof: Let $\epsilon = \frac 23$ and $N \in \Bbb R$. Let $k \in \Bbb N$ such that $k \gt N$ and suppose $m = k +1$ and $n = k$ such that $n $ is even. Then, $m,n \gt N$ and $|a_m-a_n| = |\frac {(-1)^n\;n+1}{3n}-\frac{(-1)^{n+1}\;n+2}{3(n+1)}|.$ Since $n$ is even, $m$ must be odd and so $|a_m-a_n|=|\frac{n+1}{3n}+\frac{n+2}{3(n+1)}| = |\frac{(n+1)^2+n(n+2)}{3n(n+1)}|=|\frac{2n^2+4n+1}{3n(n+1)}|\gt\frac{2n^2+4n}{3n(n+1)}=\frac{2n(n+2)}{3n(n+1)}\gt\frac{2n(n+1)}{3n(n+1)}=\frac23 = \epsilon.$ Hence, $|a_m-a_n| \ge \epsilon $ for all $m,n \gt N$ and so $\{a_n\}_n$ is not Cauchy and does not converge. $\Box$

Any help would be appreciated, thanks!

Answer 1

That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).

You could also prove that that sequence has a subsequence which converges to $\frac13$ and another one which converges to $-\frac13$. That would also prove that the sequence is not convergent.

Answer 2

You have a much simpler way to write that. Indeed, suppose that $a_n \rightarrow l$. Then $$\frac{3n}{n+1}a_n \rightarrow 3l$$

But $\frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.