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Given the sequence $a_n$ with $n \in \mathbb{N}$. $a_n = \frac{6n - 2}{4n + 1}, a = \frac{3}{2}, \epsilon := 0,01$. Show that $a$ is the limit of $a_n$. Calculate the index $n$ from which the inequality $|a_n - a| < \epsilon$ applies to all subsequent sequence elements.

I have tried to solve it like a simple limit equation to take a limit 3/2 but I guess it is the wrong.

amWhy
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Rasul Sahabadin
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    What have to tried? – JonathanZ Nov 01 '22 at 12:49
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    Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you stuck so that you receive responses that address the specific difficulties you are encountering. Please read how to ask a good question. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Nov 01 '22 at 12:49
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    I've already solved your problem. But we cannot help you until you show some thoughts or work of your own, it is a site rule. – Deepak Nov 01 '22 at 12:54
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    @RasulSahabadin I would also try to solve it like a simple limit. I think that should work. How far did you get? Did you try simplifying $|\frac{6n - 2}{4n + 1} - \frac{3}{2}|$, and see what you get? – Theo Bendit Nov 01 '22 at 13:03
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    @RasulSahabadin have you used the $\varepsilon$-$N$ definition of a limit before? If so, you just need to plug in $\varepsilon = 0.01$ to get your $N$. – Lemmon Nov 01 '22 at 13:05
  • @Lemmon its my first time of solving that type of sequence problem so i have not tried to use this:( – Rasul Sahabadin Nov 01 '22 at 13:09
  • @TheoBendit i got -(7/8n+2) – Rasul Sahabadin Nov 01 '22 at 13:10
  • @Deepak im preparing for my math exam to enter the university. and i dont know how to solve the problem so i have no solution and i tired to solve it like a simple limit equation – Rasul Sahabadin Nov 01 '22 at 13:12
  • $$ \left| {\frac{{6n - 2}}{{4n + 1}} - \frac{3}{2}} \right| = \frac{7}{{8n + 2}} < 0.01 \Longleftrightarrow \frac{{8n + 2}}{7} > 100 \Longleftrightarrow n > \frac{{349}}{4} $$ – Gary Nov 01 '22 at 13:14
  • @RasulSahabadin That's good, but I was going by the definition that Lemmon was referring to. You won't be able to follow the proof if you don't know the definition of a limit. – Theo Bendit Nov 01 '22 at 13:14

1 Answers1

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The formal definition of the limit of a sequence is the $\varepsilon$-$N$ definition. It runs:

$\lim_{n \to \infty} a_n = a$ if for any $\varepsilon > 0$, there exists some $N \in \mathbb{N}$ such that $n > N$ implies $\lvert a_n - a \rvert < \varepsilon$.

We can think of this definition as a series of "challenges". I can challenge you to provide an $N$ for $\varepsilon = 0.01$ such that every $a_n$ after $a_N$ is less than $0.01$ away from the limit of the sequence $a$.

For the limit in your question, we can see that we are given $a = \frac{3}{2}$. From this, we can see that we want

$$\lvert a_n - a\rvert =\left\lvert \frac{6n-2}{4n+1} - \frac{3}{2} \right\rvert < \varepsilon$$

$\left\lvert \frac{6n-2}{4n+1} - \frac{3}{2} \right\rvert = \frac{3}{2} - \frac{6n-2}{4n+1}$, as for all $n \in \mathbb{N}$, $\frac{3}{2} > \frac{6n-2}{4n+1}$. Thus,

$$\begin{aligned}\frac{3}{2} - \frac{6n-2}{4n+1} < \varepsilon \, &\longleftrightarrow \, -\frac{6n -2}{4n+1} < \varepsilon - \frac{3}{2}\\ &\longleftrightarrow \, \frac{3.5}{4n+1} - \frac{6n + 1.5}{4n + 1} < \varepsilon - \frac{3}{2}\\ &\longleftrightarrow \, \frac{3.5}{4n+1} < \varepsilon\\ &\longleftrightarrow \, n > \frac{\frac{3.5}{\varepsilon} - 1}{4} = \frac{7}{8\varepsilon} - \frac{1}{4}\end{aligned}$$

So, there exists an $N = \left\lfloor \frac{7}{8\varepsilon} - \frac{1}{4} \right\rfloor$ such that for any $n > N$, $\lvert a_n - a\rvert < \varepsilon$, and so $\lim_{n \to \infty}{a_n} = \frac{3}{2}$.

Now, plugging in $\varepsilon = 0.01$ to our above formula, we find that our $N = 87$, and so, for any $n > 87$, the absolute difference is less than $0.01$.

Good luck with your university entrance exam!

Lemmon
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