# Why is the constant that upper bounds every Cauchy sequence larger than the constant that bounds the Convergent sequence?

I recently read the proof of why every convergent sequence is a Cauchy sequence and the mechanics of the proof make total sense to me. However, the proof is rather different from what my intuitive brain pictured. Let me explain. For a convergent sequence we know the sequence $\{ x_n \}_n$ is converging to a specific point say $x$. What this means to me intuitively is the following picture:

In other words, since the distance from $x_n$ to $x_m$ to $x$ is bounded then its "obvious" that the distance between any $x_m$,$x_n$ should also be bounded. But not only that it seems that the distance between any $x_n$ and $x_m$ should be bounded by some $\epsilon'$ less than the original $\epsilon$ in the convergent definition. However, the proof seems to imply that it twice as big as the original $\epsilon$. Why is the relation between the epsilons the other way round from what I expect if the distance between $x_n$ to $x_m$ is included in the distance between $x_n$ and $x$. Essentially since in the picture $\| x_n - x_m \| \leq \| x_n - x \|$, then its unexpected to have the constant bounding the Cauchy sequence be twice as large.

Is the constant to loose or do I have a fundamental misconception of how the proof should have turned out?

Maybe I'll try to explain the proof in my own words and reveal my misunderstanding?

Consider a sequence $\{ x_n\}_n$ that converges to $x$. This means that $\forall \epsilon_{convergent} > 0$, $\exists N \in \mathbb{N}$ s.t. $\| x_n - x\| < \epsilon_{convergent}$. Fix a $\epsilon_{convergent}$. This corresponds to some $N$. Let it be $N_{convergent}$. Then any two points in the sequence after $N_{convergent}$ have the property $\| x_n - x\| < \epsilon_{convergent}$, $\| x_m - x\| < \epsilon_{convergent}$. Therefore consider the quantity of interest and apply the definition of norm (using triangle inequality):

$$\| x_n - x_m \| = \| x_n - x + x - x_m \| \leq \| x_n - x\| + \| x_m - x \|$$

but since we know both are bounded by $\epsilon_{convergent}$ we get that for that same $N_{convergent}$ the above difference is bounded by $\epsilon_{cauchy} = 2 \epsilon_{convergent}$. Which we used all the rules and axioms correctly so mechanically it looks fine but the constant $\epsilon_{cauchy}$ seems to be larger than I expected (when I thought it should have obviously been smaller), which could mean my intuition is wrong. I just hope to correct it or maybe I don't understand the proof as well as I thought.

• In general you do not have $\Vert x_n - x_m \Vert \leq \Vert x_n -x \Vert$. Just draw your picture slightly different, namely let $(x_n)_{n\geq 1}$ approach $x$ not only from the left, but from both sides. This is exactly where the factor 2 is coming from. – Severin Schraven Sep 24 '16 at 22:12
• @SeverinSchraven I see that you have an argument that explains why I'm wrong but I don't understand what I'm suppose to draw. I know you described it but I don't get it I guess. – Charlie Parker Sep 24 '16 at 22:17
• @SeverinSchraven: good answer. I suggest you post it as an answer and not as a comment. – Martin Argerami Sep 24 '16 at 22:18

In general you do not have $\Vert x_n−x_m\Vert \leq \Vert x_n−x \Vert$. Just draw your picture slightly different, namely let $(x_n)_{n\geq 1}$ approach x not only from the left, but from both sides. This is exactly where the factor 2 is coming from.

Look for example at the sequence $-1, 1, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{3}, \frac{1}{3}\dots$ (or formally correct $x_{2n+1}:= -\frac{1}{n}$ and $x_{2n+2}:=\frac{1}{n}$).

This sequence converges to $x=0$. We have

$$\Vert x_{2n+ 1} - x_{2n+2} \Vert = \Vert -\frac{1}{n} - \frac{1}{n} \Vert = \frac{2}{n},$$

but

$$\Vert x_{2n+1} - x \Vert = \Vert -\frac{1}{n} - 0 \Vert = \frac{1}{n}.$$

• a picture would make this answer perfect. Just a suggestion. Thanks btw. :) – Charlie Parker Sep 25 '16 at 16:50

take the example of a sequence which converges to x alternatively from the left and the right sides like

$$x_n=sin(n)e^{-n}$$

then you could have

$$|x_n-x_m|>|x_n-x|$$