<s>[INST] Show that the sequence $a_n = \frac{(n+1)^2 -n^2}{n}$ converges and give its limit. can somebody tell me if I did this somewhat correctly? First I estimated the limit of the sequence by calculating the first few results of the sequence and it looks like it is converging towards $2$. Using the Cauchy criterion we then have: $$\forall \epsilon > 0 \exists N\in\mathbb{N}_0 \forall n \geq N: \left|\frac{(n+1)^2 -n^2}{n} - 2\right| < \epsilon$$
and after doing a bit of math we have $$\left|\frac{(n+1)^2 -n^2}{n} - 2\right| < \epsilon \Leftrightarrow \left|\frac{1}{n}+2-2\right| < \epsilon \Leftrightarrow \frac{1}{n} < \epsilon \Leftrightarrow \frac{1}{\epsilon} < n $$ which means that we need a $N\in\mathbb{N}$ with $N > \frac{1}{\epsilon}$ and we have $\forall n\geq N$:$$\left|\frac{(n+1)^2 -n^2}{n} - 2\right| = \dots=\frac{1}{n}\leq\frac{1}{N}<\epsilon\,.$$
And hence our sequence $a_n \longrightarrow a_\infty$ with $a_\infty = 2$$._{\,\,\square}$

Did I do this right?
 [/INST] To do this more cleanly but also rigorously, note that given $\epsilon > 0$, it is possible to find $N$ such that $N > \dfrac{1}{\epsilon}$.  Then for all $n > N$, we have that $\dfrac{1}{n} < \epsilon$.  From here, we can achieve the desired result:
$$
\epsilon > |{\dfrac{1}{n}}|  = |\dfrac{2n + 1 - 2n}{n}| = |\dfrac{2n + 1}{n} - 2 | = |\dfrac{n^2 + 2n + 1 - n^2}{n} - 2 | = |\dfrac{(n + 1)^2 - n^2}{n} - 2|
$$
Thus, given any $\epsilon > 0$, if we choose any $n > N$, we have that 
$$
|\dfrac{(n + 1)^2 - n^2}{n} - 2| < \epsilon
$$
Since such $N$ exists, this implies that the sequence $a_n$ converges to 2.
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