An equivalent expression of Cauchy Criterion?

Question

For a sequence $\{a_n\}$, if $$ \forall \epsilon>0 \ \exists N>0, \forall k \in \mathbf{N}, \ |a_{N+k}-a_N|<\epsilon \ $$ Then $\{a_n\}$ converges and hence is a Cauchy sequence.

Now how about changing the inequality above to $|a_{N+k}-a_N|< a_{N+k}\cdot \epsilon$, or equivalently $|\frac{a_N}{a_{N+k}} - 1| < \epsilon$? Does the sequence still converge?

Answer 1

Does the sequence still converge?

Yes. Assume the criterion in the question holds and, for every $n\geqslant1$, let $N_n$ such that, for every $j\geqslant i\geqslant N_n$, $|a_i-a_j|\leqslant a_j/2^n$. Then, for every $j\geqslant N_1$, $\frac23a_{N_1}\leqslant a_j\leqslant2a_{N_1}$, hence $(a_k)$ is bounded. Furthermore, for every $n\geqslant1$, for every $j\geqslant i\geqslant N_n$, $a_i\leqslant(1+2^{-n}) a_j$ and $(1-2^{-n})a_j\leqslant a_i$ hence $(1-2^{-n})\limsup a_k\leqslant\liminf a_k$. Thus, $(a_k)$ converges.

Note that the condition $$ \forall \epsilon>0, \ \exists N>0, \forall k \geqslant0, \ |a_{N+k}-a_N|<a_N\cdot\epsilon, $$ also implies convergence (the proof is similar).

Answer 2

You second proposition is actually only defined if the sequence is always non-zero after some $n$. Otherwise, if it takes value $0$ infinitely often, then you cannot even write the criterion you suggest.

As André Nicolas pointed out, for the first one, take $0 < \alpha < 1$, and set $a_n\stackrel{\rm{}def}{=} a^n$. It does converge, but your criterion would imply that $$ \forall k\in\mathbb{N},\quad |a^k-1|\leq a^k\varepsilon $$ the LHS goes to 1, the RHS to $0$. So your criterion would not be even allow you to conclude for rather elementary sequences...

In both cases, regardless of whatever it still implies convergence, the criterion would not be a very good tool to use.