<s>[INST] Prove sequence $\left(\frac{1}{6n^2+1}\right)$ converges to $0$ I am asked to verify that the sequence $\left(\frac{1}{6n^2+1}\right)$ converges to $0$:
$$\lim \frac{1}{6n^2+1}=0.$$  
Here is my work:
$$\left|\frac{1}{6n^2+1}-0\right|<\epsilon$$
$\frac{1}{6n^2+1}<\epsilon$, since $\frac{1}{6n^2+1}$ is positive
$$\frac{1}{\epsilon}<6n^2+1$$
$$\frac{1}{\epsilon}-1<6n^2$$
$$\frac{1}{6\epsilon}-\frac{1}{6}<n^2$$
At this point, I am stuck. I'm not sure if I take the square root of both sides if I then have to deal with $\pm\sqrt{\frac{1}{6\epsilon}-\frac{1}{6}}$. That doesn't seem right. 
The book provides the answer:
$$\sqrt{\frac{1}{6\epsilon}}<n$$
But I don't understand (1) what happened to the $\frac{1}{6}$, and (2) why there's not a +/- in front of the square root. 
Any help is greatly appreciated. 
 [/INST] Let $\epsilon > 0$. Let's sketch it first. $$\frac{1}{6n^2+1} < \frac{1}{6n^2} < \epsilon.$$ Look: $$\frac{1}{6n^2} < \epsilon \iff 1 < 6n^2\epsilon \iff \frac{1}{6\epsilon} < n^2 \iff n > \frac{1}{\sqrt{6\epsilon}}.$$

Now we begin. Let $\epsilon > 0$. Then exists $n_0 \in \Bbb N$ such that $n_0 > 1/\sqrt{6\epsilon}$. If $n \geq n_0$, then: $$\left|\frac{1}{6n^2+1}\right| = \frac{1}{6n^2+1} < \frac{1}{6n^2} \leq \frac{1}{6n_0^2} < \epsilon.$$
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