# Proof explanation - Sequence convergence

In my lecture notes there is a side note to the proof for the example below that the last inequality $$\frac{14}{n}\lt{\epsilon}$$ in the equation is not always true and only holds under the condition that $$n\gt \frac{14}{\epsilon}$$, hence the sentence continues with the condition "provided $$n\gt \frac{14}{\epsilon}$$".

I'm confused and can't see this point. Can someone please explain to me when this does not hold, so that we proceed with the following sentence. After finding a suitable inequality for $$n_0$$ in other examples I usually just proceed with "take $$n_0 \in \mathbb{N}$$, with $$n_0\gt ...$$ ".

Example:

Let $$(x_n)_{n=1}^{\infty}$$ be given by $$x_n=\frac{3n-2}{n+4}.$$ show that $$x_n\rightarrow{3}$$ as $$n\rightarrow{\infty}$$, directly from the definition.

Solution:

Let $$\epsilon \gt{0}$$. For $$n\in \mathbb{N}$$, we have $$\left\lvert {\frac{3n-2}{n+4}-3} \right\rvert =\left\lvert{\frac{3n-2-3(n+4)}{n+4}}\right\rvert=\frac{14}{n+4}\lt\frac{14}{n}\lt{\epsilon},$$

provided $$n\gt\frac{14}{\epsilon}$$. Take $$n_0\in \mathbb{N}$$ with $$n_0\geq\frac{14}{\epsilon}$$. Then for $$n\in \mathbb{N}$$ with $$n\geq n_0$$ we have $$|x_n -3|\lt3$$, so that $$x_n\rightarrow3$$ as $$n\rightarrow\infty$$.

## 2 Answers

There is no reason why the inequality $$\frac{14}n<\varepsilon$$ would hold in general. If, for instance, $$\varepsilon=n=1$$, then it does not hold. Whoever wrote that is actually just saying that$$\frac{14}n<\varepsilon\iff n>\frac{14}\varepsilon,$$which is clearly true.

• OMG!Our answers are actually the same. Even the example!! – Thomas Shelby Feb 14 at 18:04
• @ThomasShelby Well, after all it was the natural example in this context. – José Carlos Santos Feb 14 at 18:06
• So would it be ok to proceed with "take $n_0 \in \mathbb{N}$ with $n_0 \gt \frac{14}{\epsilon}$ so that for $n\in \mathbb{N}$ with $n\geq{n_0}$ we have that $|x_n - 3|\lt{\epsilon}$. Therefore $x_n\rightarrow{3}$ as $n\rightarrow{\infty}$" – Adnaan Feb 14 at 18:07
• No, it would not be OK. Pick $n_0$ such that $n_0>\frac{14}\varepsilon$ instead. – José Carlos Santos Feb 14 at 18:08

Let's take $$\varepsilon =1$$. Then note that $$\frac {14}1\gt1$$.So we don't have $$\frac{14}n\lt 1$$ for all $$n\in \Bbb N$$. But, if you take $$n\gt 14$$, we have $$\frac {14}n\lt1$$. So in general, $$\frac {14}n\lt\varepsilon \iff n\gt\frac {14}{\varepsilon}$$.