Using Cauchy convergence to prove that $\sum_{k=1}^{n} z_{k}$ converges if $\sum_{k=1}^{n}\left | z_{k} \right |$ does

Question

Consider the sequence $\left \{ z_{k} \right \}$ in $\mathbb{R}.$  Let the sequence  $$\left \{ x_{n} \right \} = \sum_{k=1}^{n}z_{k}\ \ \ \   \text{and}\ \ \  \left \{ y_{n} \right \} = \sum_{k=1}^{n}\left | z_{k} \right |.$$ My goal is to show that if  $\left \{ y_{n} \right \}$ converges, then $\left \{ x_{n} \right \}$ converges, using the ideas of a Cauchy sequence and  $\textbf{without}$ any ideas from infinite series.  

I know that a sequence $\left \{ a_{n} \right \}$ in $\mathbb{R}$ is a Cauchy sequence if$ \ $ $\forall \ \epsilon > 0, \exists \ N \in \mathbb{N}$ such that whenever natural numbers $m,n > N$,$$\left | a_{m}-a_{n} \right |< \epsilon.$$

From what I understand, this is saying that the distance between any two elements of $\left \{ a_{n} \right \}$ get closer and closer to each other as $n\rightarrow \infty.$  It is sort of like the elements are "bundling up" near the supremum.  

One thought I had initially was to consider one sequence as a subsequence of the other, but I do not think that would be true in this case.  

Does anyone have any ideas?  Thanks in advance! 

Answer 1

Take any $\epsilon>0$. You assume the sequence $y_n$ is a cauchy sequence so you know that:

$\exists (n_0\in\mathbb{N})\forall(m,n\geq n_0)[|y_m-y_n|<\epsilon]$

Well, so let's take $m>n\geq n_0$. Then using the triangle inequality we get:

$|x_m-x_n|=|\sum_{k=1}^{m}z_{k}-\sum_{k=1}^{n}z_{k}|=|\sum_{k=n+1}^{m}z_{k}|\leq \sum_{k=n+1}^{m}|z_{k}|=\sum_{k=1}^{m}|z_{k}|-\sum_{k=1}^{n}|z_{k}|=y_m-y_n<\epsilon$

So $x_n$ is also a cauchy sequence.

Answer 2

For $m<n$ You have $$|x_n-x_m|=|\sum_{k=m+1}^{n}z_k|\leq\sum_{k=m+1}^{n}|z_k|=|y_n-y_m|$$ by the triangle inequality. Can You take it from here?

Answer 3

Concerning your idea: no, it is not true in general.

Take $\varepsilon>0$. Since $(y_n)_{n\in\mathbb N}$ converges, there is a natural $N$ such that$$m,n\geqslant N\implies\lvert y_m-y_n\rvert<\varepsilon.\tag1$$Now, note that, in$(1)$, if $m=n$, then the inequality trivially holds. So, nothing is changed if we add to $(1)$ that $m\neq n$. So, in fact, $(1)$ is equivalent to$$m>n\geqslant N\implies\lvert y_m-y_n\rvert<\varepsilon.$$But this is equivalent to$$m>n\geqslant N\implies\sum_{k=n+1}^m\lvert z_k\rvert<\varepsilon.$$Therefore, if $m,n\in\mathbb N$ and $m>n\geqslant N$, we have$$\left\lvert\sum_{k=n+1}^mz_k\right\rvert\leqslant\sum_{k=n+1}^m\lvert z_k\rvert<\varepsilon$$It follows that $(x_n)_{n\in\mathbb N}$ is a Cauchy sequence and therefore that it converges.