What is the mistake in my proof of limit of a sequence? I'm completely confused by the formal proofs of limits of sequences using the ε-N definition.
For example, lets take the sequence a = (2n)/(n+1) where the limit is 2.
For proof, I would solve the inequality |(2n)/(n+1) - 2|<ε and get n> (2/ε)-1.
That means that there exists an N > (2/ε)-1, such that for any n ≥ N, |(2n)/(n+1) - 2|<ε is true.
However, lets assume the limit is 1. Than we can do basically the same thing;
We can solve the inequality |(2n)/(n+1) - 1|<ε and get n > (1-ε)/(1+ε), which means that there exist an N > (1-ε)/(1+ε), such that for every n ≥ N, (2n)/(n+1) - 2|<ε is true.
What I want to know is what is the difference between the first procedure and the second. I believe it is possible that I have completely misunderstood how these proofs work, since this is the first time I've been introduced to them. So please explain thoroughly what is wrong in the second part and what is it that I don't get.
 A: Hmm, two things:
(i) You don't need to solve the inequality exactly, that will work if you can do it, but it is usually more work than is necessary.
(ii) You need to show that for any $\epsilon>0$ there is some $N$ such that for all $n \ge N$ then $|x_n -L| < \epsilon$. 
If $x_n = {2n \over n+1}$ and $L=1$, then a little work shows that if $n \ge 2$ then
$|x_n-L| \ge {1\over 3}$. So it cannot converge to $L$.
A: You evaluated the second wrongly. For every positive integer $n$ and $0 < \epsilon < 1$,
$$
\begin{align}
& \left| {2n \over n+1} - 1 \right| = \left| {n -1 \over n+1} \right| & < &  \epsilon \\
\iff & {n -1 \over n+1} & < & \epsilon \\
\iff & n-1 & < & \epsilon \left( n + 1\right) \\
\iff & n (1 - \epsilon) & < & \epsilon + 1 \\
\iff & n & < & {1 + \epsilon \over 1 - \epsilon}
\end{align}
$$
A: The limit can't be $1$, because for $\epsilon = \frac{1}{10}$ conclusion fails. Here is the proof:
If $\left | \frac{2n}{n+1}- 1\right | < \frac{1}{10}$
$ \implies  n < \frac{11}{9} $
so, for $n \ge 2$, above conclusion does not hold. To prove that limit is $2$ (which comes by intuition), you need to prove the result is true for every $\epsilon$. But if your intuition say that limit can't be 1, then try to find a single $\epsilon$.
A: If $|\frac{2n}{n+1}-2|<\varepsilon$ then try
$|\frac{2n}{n+1}-2|=\varepsilon$ to see what happens at the border
of the $\varepsilon$ neighborhood of the limit $2$,
then 
$|2n-2(n+1)|=\varepsilon(n+1)$ so you have
$\varepsilon(n+1)=2$
or 
$$\varepsilon=\frac{2}{n+1}.$$
This indicates how big $n$ must be to achieve the given estimation of a specific $\varepsilon$.
Addendo 
Could you do silmilar for
$$\left|\frac{2n}{n+1}-1\right|=\varepsilon?$$
This is asking for the $\varepsilon$-distance between the steps $a_n$ and the point at $1$.
This leads you to $|n-1|=\varepsilon(n+1)$. So $n-1=\varepsilon n+\varepsilon$ for $n>0$, then
$$n=\frac{1+\varepsilon}{1-\varepsilon}.$$
The behavior for this function in the range $0\le\varepsilon<1$
is strictly monotonic increasing, defined at $\varepsilon$ equals zero giving $n=1$.
So, the steps $a_1=0.66...$ and $a_2=1.33...$ are the closers to $1$ and the remaining $a_3,a_4,a_5,...$ are going the way to $2$.
