Can a proof that a sequence is Cauchy depend on the difference between sequence indices?

Question

So, the definition of a Cauchy sequence is:

For any $\epsilon > 0$ there exists $M > 0$ such that for all $m, n > M$, $\|x_m - x_n\| < \epsilon$.

My question is, is it valid for the choice of $M$ to depend on $m-n$?

Specifically, I have that $\lim_{k \rightarrow \infty} \|x_k\| = 0$. I need to prove that the sequence $S_n = \sum_{k=1}^n x_n$ is Cauchy.

I first assume that $m > n$ and define $l = m-n$. Then let $M$ be the number such that $$k>M \implies \|x_k\| < \frac{\epsilon}{l}.$$

Then if $n,m>M$ $$\|S_m - S_n\| = \|x_{n+1} + x_{n+2} + \cdots x_{n+l}\| \leq \|x_{n+1}\| + \cdots \|x_{n+1}\| < \frac{\epsilon}{l} + \cdots + \frac{\epsilon}{l} = \epsilon.$$

I have a strong feeling that this argument isn't valid, but I can't think of any other way of proving it.

Answer 1

No, this isn't valid. In the definition of a Cauchy sequence, "there exists $M>0$" comes before "for all $m,n>M$", so you have to choose $M$ before you know anything about $m$ and $n$.

(By the way, the result you are trying to prove is not true, which is probably the reason you are having difficulty proving it. For instance, think about what happens when your space is $\mathbb{R}$ with the usual norm and $x_k=1/k$.)

Answer 2

You have been provided that $m, n \geq M$, however, you know nothing of their relation ($m< n, m> n, m=n$?).
It's a corollary in analysis: If a sequence $x_n\to L\in\mathbb{R}$, then it is Cauchy.

Assume $x_n\to L\in\mathbb{R}$. Fix $\varepsilon >0$, then there exists index $P\in\mathbb{N}$ such that
$n\geq P\Rightarrow |x_n-L|<\frac{\varepsilon}{2}$ from which we have that
$m,n\geq P\Rightarrow |x_n-L-x_m+L|\leq |x_n-L|+|L-x_m| = |x_n-L|+|x_m-L|< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon$
therefore $x_n$ is Cauchy.

All you really have to do is show $S_n$ converges, but does it?