Series convergence and limit I am trying to solve an exercise that asks me this:
Prove using the $ε–N$ method that the sequence
$a(n) = \frac{n^2 + n - 1}{n^2 + n}$
converges and state the limit.
My attempt is the following:
First, I guess that the $\lim \limits_{n\to\infty} a(n) = 1$.
Then we require 
\begin{equation}\begin{split}
|a(n)-1| < ε &\iff \left| \frac{n^2 + n - 1}{n^2 + n} - 1 \right| < ε \\
&\iff \frac{1}{n^2 + n} < ε \\
&\iff \frac{1 }{ ε} < n^2 + n \\
&\iff n^2 + n > \frac{1}{ε}
\end{split}\end{equation}
Now, we take another number $2n^2$ which is greater or equal than $n^2 + n$ and say:
$2n^2 \geq n^2 + n > \frac{1}{\epsilon}$
$\implies 2n^2 > \frac{1}{\epsilon}$
$\implies n > \frac{1}{\sqrt{2ε}}$
Therefore $N(ε) = \left\lceil\frac{1}{\sqrt{2\epsilon}}\right\rceil$
Is this correct? Can we just take another number greater than or equal to $n^2 + n$ and just continue to find the solution?
Any help is welcomed. Thanks! :)

$2n^2 > 1/\epsilon$ will not guarantee $n^2+n > 1/\epsilon$.
You need to find a nice function of $n$, that lies between $n^2+n$ and $1/\epsilon$, which will allow you to solve for $n$ in terms of $\epsilon$. We can in fact choose $n>1/\epsilon$, since this will ensure that $n^2+n > 1/\epsilon$.
