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 $\displaystyle \frac{n+1}{2n}$ is $\displaystyle \frac{1}{2}$. I understand that the first step is to set $\displaystyle \frac{n+1}{2n} - \frac{1}{2} < \epsilon$. I then solve for n and get $\displaystyle n>\frac{1}{2\epsilon}$. So I think I understand this tells me that for any $\displaystyle \epsilon$, picking an $\displaystyle n>\frac{1}{2\epsilon}$ will give a value less than $\displaystyle \epsilon$. I'm just not sure where to go next with the proof. I've arbitraily tried different values than $\displaystyle \frac{1}{2}$ for the limit and every time I can solve $\displaystyle \frac{n+1}{2n} -$ arbitrary number for $\displaystyle n > $ something with $\displaystyle \epsilon$ 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 $\displaystyle n>\frac{1}{2\epsilon}$ and somehow work in the "other direction" to prove that indeed any $\displaystyle n>\frac{1}{2\epsilon}$ works, but I've been unable to follow. Thanks for any help.

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 $\displaystyle \epsilon$ then I pick an N. I win if given my N for every n > N, $\displaystyle |a_{n} - L| < \epsilon$

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

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

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

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

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

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

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

Q.E.D.

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

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