Thread: Proving Limit of a Sequence using Epsilon N

1. Proving Limit of a Sequence using Epsilon N

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.

2. Re: Proving Limit of a Sequence using Epsilon N

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, $|a_{n} - L| < \epsilon$

3. Re: Proving Limit of a Sequence using Epsilon N

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

4. Re: Proving Limit of a Sequence using Epsilon N

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.

5. Re: Proving Limit of a Sequence using Epsilon N

\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*}.

6. Re: Proving Limit of a Sequence using Epsilon N

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

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