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?
 A: Note that $$a_n = \dfrac{(n+1)^2-n^2}{n} = \dfrac{(n^2+2n+1)-n^2}{n} = 2 + \dfrac{1}{n}\,.$$ Therefore, $$\lim a_n = \lim {}\biggl( 2 + \dfrac{1}{n} \biggr) = \lim 2 + \lim \dfrac{1}{n} = 2 + 0 = 2.$$ 
A: Yes it's right, infact:
$$\frac{(n+1)^2 -n^2}{n}=\frac{n^2+2n+1-n^2}{n}=\frac{2n+1}{n}\to 2$$
A: Mark's method is cleaner, but yours is nice and rigorous. I do not see any issues with your proof. You did this right!
A: 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.
A: $$\frac{(n+1)^2 -n^2}{n}-2=\frac{n^2+2n+1-n^2}{n}-2=\frac{2n+1}{n}-2=\frac{1}{n}
\to 0$$
A: There is only a minor error in what you wrote. That's when you state: “Using the Cauchy criterion we then have”. No, you did not use the Cauchy criterion. What you did use was the definition of convergent sequence.
