Proving Limit of a Sequence using Epsilon N

**KarlKarlJohn** · Nov 5th 2012, 07:17 PM

I am having a lot of trouble with the concept of proving the limit of a sequence using epsilon n. As an example, I am working on trying to prove that the limit of the sequnce $\frac{n+1}{2n}$ is $\frac{1}{2}$ . I understand that the first step is to set $\frac{n+1}{2n} - \frac{1}{2} < \epsilon$ . I then solve for n and get $n>\frac{1}{2\epsilon}$ . So I think I understand this tells me that for any $\epsilon$ , picking an $n>\frac{1}{2\epsilon}$ will give a value less than $\epsilon$ . I'm just not sure where to go next with the proof. I've arbitraily tried different values than $\frac{1}{2}$ for the limit and every time I can solve $\frac{n+1}{2n} -$ arbitrary number for $n >$ something with $\epsilon$ in it. So I must be doing something wrong as it seems using my method I can prove the limit to be any arbitrary number. In searching online and other forums it seems there is maybe a last step where I take my $n>\frac{1}{2\epsilon}$ and somehow work in the "other direction" to prove that indeed any $n>\frac{1}{2\epsilon}$ works, but I've been unable to follow. Thanks for any help.

**KarlKarlJohn** · Nov 5th 2012, 07:29 PM

As another note, the way the concept was explained in class was that the limit L exists if I can always beat you at a game where you pick an $\epsilon$ then I pick an N. I win if given my N for every n > N, $|a_{n} - L| < \epsilon$

**Prove It** · Nov 5th 2012, 08:02 PM

To prove the limit of the sequence $\displaystyle \begin{align*} \frac{n+1}{2n} \end{align*}$ is $\displaystyle \begin{align*} \frac{1}{2} \end{align*}$ , you need to prove $\displaystyle \begin{align*} \lim_{n \to \infty}\frac{n +1}{2n} = \frac{1}{2} \end{align*}$ by showing $\displaystyle \begin{align*} n > M \implies \left| \frac{n + 1}{2n} - \frac{1}{2} \right| < \epsilon \end{align*}$ . Working on the second inequality we have

$\displaystyle \begin{align*} \left| \frac{n + 1}{2n} - \frac{1}{2} \right| &< \epsilon \\ \left| \frac{n + 1 - n}{2n} \right| &< \epsilon \\ \left| \frac{1}{2n} \right| &< \epsilon \\ \frac{1}{2|n|} &< \epsilon \\ 2|n| &> \frac{1}{\epsilon} \\ |n| &> \frac{1}{2\epsilon} \end{align*}$

So choose $\displaystyle \begin{align*} M = \frac{1}{2\epsilon} \end{align*}$ and reverse each step and you will have your proof

**KarlKarlJohn** · Nov 5th 2012, 09:50 PM

Thanks for the quick reply! I fully understand everything right up to the last part.

So choose $\displaystyle \begin{align*} M = \frac{1}{2\epsilon} \end{align*}$ and reverse each step and you will have your proof

This is the part that trips me up. so when I say $M = \frac{1}{2\epsilon}$ I just plug that straight back into the right side of $n > \frac{1}{2\epsilon}$ and go back to the beginning? I guess I'm have trouble understanding how that proves $\frac{1}{2}$ is the limit. For example, if I instead begin with $\displaystyle \begin{align*} n > M \implies \left| \frac{n + 1}{2n} - \frac{1}{4} \right| < \epsilon \end{align*}$ I can get down to $n > \frac{2}{4\epsilon -1}$ and also reverse my steps, but $\frac{1}{4}$ isn't the limit.

**Prove It** · Nov 5th 2012, 10:19 PM

$\displaystyle \begin{align*} n &> \frac{1}{2\epsilon} \\ |n| &> \frac{1}{2\epsilon} \textrm{ since } n > 0 \implies n = |n| \\ \frac{1}{|n|} &< 2\epsilon \\ \frac{1}{2|n|} &< \epsilon \\ \left| \frac{1}{2n} \right| &< \epsilon \\ \left| \frac{n + 1 - n}{2n} \right| &< \epsilon \\ \left| \frac{n + 1}{2n} - \frac{n}{2n} \right| &< \epsilon \\ \left| \frac{n + 1}{2n} - \frac{1}{2} \right| &< \epsilon \end{align*}$

Q.E.D.

The reason it works is because you need to show there exists some value $\displaystyle \begin{align*} M \end{align*}$ such that no matter how big it is, your function value is always within a band of width $\displaystyle \begin{align*} \epsilon \end{align*}$ from the limit value. In symbols, that means $\displaystyle \begin{align*} \lim_{n \to \infty} f(n) = L \end{align*}$ if $\displaystyle \begin{align*} x > M \implies \left| f(n) - L \right| < \epsilon \end{align*}$ .

**KarlKarlJohn** · Nov 6th 2012, 10:17 AM

Ok, great, that clears it up for me. Thanks for all the help!

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