Formally prove that $\lim(\frac{2n}{n+1})=2$

Question

Taking Real Analysis for the first time and having trouble grasping what i need to show for the proof. I know i need for any $\epsilon > 0$, there exists a positive integer $N$, such that if $n \geq N$, then $$\biggl|\frac{2n}{n+1}-2\biggr|= \frac{2}{n+1} < \epsilon $$

I'm not sure where to go from here. Do I need to solve for $n$ on the right hand side?

Okay then i understand nothing on what to do. Whats the first step? I understand how that doesn't make any sense. — Ringomain, Sep 14 '18 at 23:09

score 2 · Answer 1 · answered Sep 14 '18 at 23:12

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Notice first that you have a calculation mistake, it should be

$$\frac{2n}{n+1}-2 = \frac{2n}{n+1}-\frac{2(n+1)}{n+1} = -\frac{2}{n+1}$$

Now, this means that

$$\left|\frac{2n}{n+1}-2\right| = \frac{2}{n+1}$$

So it is left to show that $\lim_{n\rightarrow\infty}\frac{2}{n+1}=0$. You should slightly modify the proof that $\lim_{n\rightarrow\infty}\frac{1}{n}=0$ see here

answered Sep 14 '18 at 23:12

Yanko

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I think OP wants an $\epsilon,\delta$ proof. – Ahmad Bazzi Sep 14 '18 at 23:16
@AhmadBazzi I know but, we talk about sequences not functions. Anyway, he should check out the link I provided, there's a very detailed proof. – Yanko Sep 14 '18 at 23:17
Yes i saw it too. You convinced me now because it is an epsilon proof. +1 – Ahmad Bazzi Sep 14 '18 at 23:21

score 1 · Answer 2 · answered Sep 14 '18 at 23:15

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Hint: Given that $$\left |\frac{2n}{n+1}-2\right|=\left |\frac{2n-2n-2}{n+1}\right|=\frac{2}{n+1}<\epsilon$$ What inequality must $n$ satisfy for this to be true? What could you choose as $N$?

answered Sep 14 '18 at 23:15

cansomeonehelpmeout

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$n > \epsilon?$ – Ringomain Sep 14 '18 at 23:25
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@Ringomain $\epsilon$ is close to $0$, so yes, $n$ needs to be greater than $0$, but that's no enough. If you look at $$\frac{2}{n+1}<\epsilon$$ and manipulate the inequality you get $$n>\frac{2}{\epsilon}-1$$, so if you choose an integer $N$ such that $$N>\frac{2}{\epsilon}-1$$ you're done! – cansomeonehelpmeout Sep 14 '18 at 23:30

score 1 · Answer 3 · answered Sep 14 '18 at 23:16

Given $\epsilon >0$. By the Archimedean's property there is an $N\in \mathbb{N}$ such that $\frac{2}{N}<\epsilon$.

Now let $m\geq N$. Notice that $|\frac{2m}{m+1}-2|=|\frac{2m}{m+1}-\frac{2m+2}{m+1}|=\frac{2}{m+1}<\frac{2}{m}\leq \frac{2}{N}<\epsilon$.

Hence the the limit is 2.

score 1 · Answer 4 · answered Sep 14 '18 at 23:19

As an alternative note that

$$\lim_{n\to \infty} \frac{2n}{n+1}=\lim_{n\to \infty} \frac{2n+2-2}{n+1}=\lim_{n\to \infty} \left(2-\frac{2}{n+1}\right)=2-\lim_{n\to \infty} \left(\frac{2}{n+1}\right)$$

then it suffices to prove that $\lim_{n\to \infty} \left(\frac{2}{n+1}\right)=0$ which indeed corresponds to your step that is

$$\frac{2}{n+1} < \epsilon \implies n>\frac 2 \epsilon -1$$

score 1 · Answer 5 · answered Sep 14 '18 at 23:20

Let $\varepsilon>0$ be given. By the Archimedean Property, we may find a positive integer $N$ so that $\varepsilon N>2$. Thus if $n \geq N$, then we have \begin{aligned} \left|\frac{2n}{n+1}-2\right|&=\left|\frac{2n}{n+1}-\frac{2(n+1)}{n+1}\right| \\& =\frac{2}{n+1} \\& \leq \frac{2}{n}<\varepsilon .\end{aligned} Therefore, $\lim_{n \to \infty} \frac{2n}{2n+1}=2$.

Formally prove that $\lim(\frac{2n}{n+1})=2$

5 Answers5