<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] $$\frac{(n+1)^2 -n^2}{n}-2=\frac{n^2+2n+1-n^2}{n}-2=\frac{2n+1}{n}-2=\frac{1}{n}
\to 0$$
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