Proof that a sequence defines a Cauchy sequence

Question

I recently did a question that said to prove the sequence ${\{x_n\}}=\frac{3n}{2n+1}$ is Cauchy and I'm unsure on if my proof is valid.

First I stated that ${\{x_n\}}$ is Cauchy $\iff \forall \epsilon > 0, \exists N \in \mathbb{N}$ s.t: $|x_n-x_m| < \epsilon, \forall m,n \ge N$

I then did the following calculations: Consider the quantity $|x_n-x_m|=\left|\frac{3n}{2n+1}-\frac{3m}{2m+1}\right| = \left|\frac{3n}{(2n+1)(2m+1)} + \frac{-3m}{(2n+1)(2m+1)}\right| \le \frac{3n}{(2n+1)(2m+1)} + \frac{3m}{(2m+1)(2n+1)} < \frac{3}{2m+1} + \frac{3}{2n+1}$

I then stated that we wanted to choose a value for $N$ such that $|x_n-x_m| < \frac{3}{2m+1}+\frac{3}{2n+1} < \epsilon + \epsilon = 2\epsilon$

Therefore, if we let $\frac{3}{2m+1}<\epsilon$, we can rearrange to obtain $m > \frac{1}{2}\left(\frac{3}{\epsilon}\right)-1$

Therefore, choose $N=\left\lceil\frac{1}{2}\left(\frac{3}{\epsilon}\right)-1\right\rceil\square$

I'm very new to real analysis so I don't have a full grasp on the required levels of formality just yet, so any tips or pointers would be appreciated :)

Answer 1

The proof looks fine, but I suggest beginners follow the definition pedantically. The statement to show, as you stated, is: $$ \forall \varepsilon>0\ \exists N\in\mathbb{N}\ \forall m,n>N:|x_n-x_m|<\varepsilon $$ If we strictly follow this definition, we'll notice that any proof should end with $|x_n-x_m|<\varepsilon$. What you have written, strictly speaking, is only a draft which helps you find the $N$. You didn't show us that this choice of $N$ actually works (you demonstrated a motivation why it would work). In fact, your $N$ could be zero if $\varepsilon$ is large enough, so you probably cannot turn your draft into the actual proof directly.

For example, you could write something like this:

Let $\varepsilon>0$. Choose $N=\lceil \frac{3}{2\varepsilon}-1 \rceil$. Note that $N\ge 0$, so $N\in\mathbb{N}$. Let $m,n>N$. Then ... Therefore, $|x_n-x_m|<\varepsilon$.

Another thing is that you can choose a simpler $N$ (not necessarily an explicit expression). Note that if $m>N>0$, then $$ \frac{3}{2m+1}<\frac{3}{m}<\frac{3}{N}. $$ You would get a simpler explicit expression for $N$ if you did a little more simplification of the inequality like above. However, you can avoid an expression of $N$ at all by saying that "by Archimeandean principle, there exists $N\in\mathbb{N}$ such that $\frac{1}{N}<\frac{\varepsilon}{6}$." The expression does not actually matter in this case because it is only used to guarantee that $\frac{3}{N}<\frac{\varepsilon}{2}$, so you don't need to construct it explicitly.

Hope this helps!