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. 
 A: Note that $\left| {1 \over 6n^2 + 1} - 0 \right| < \left| 1 \over 6n^2 \right|$. Hence if we can bound the larger term by $\epsilon$ you're done. I.e., given an arbitrary $\epsilon > 0$, we can choose a lower bound $N$ by
$$N = \sqrt{\frac{1}{6\epsilon}}$$
Then
$$n > N \Rightarrow \left| {1 \over 6n^2 + 1} - 0 \right| < \left| 1 \over 6n^2 \right| <  {1 \over 6N^2} = \epsilon$$
A: 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: $$\left|\frac{1}{6n^2+1}\right| = \frac{1}{6n^2+1} < \frac{1}{6n^2} \leq \frac{1}{6n_0^2} < \epsilon.$$
A: Hy!
I will take a similar approach to that of Simon's and that's to find a "similar" N for which numbers greater than him are smaller than an upper bound defined in N(that's our epsilon).
It's actually a convergence test done with comparison.
$$\text{Given we have two sequences, } a_n \text{ and } b_n \text{ if} \\ \sum_0^na_n \to convergent \quad \text{than} \quad \sum_0^nb_n \to convergent \quad if\\$$ 
$$\lim_{x\to\infty} \dfrac{a_n}{b_n} = c, \quad where   \quad c\in(0, \infty)$$ 
That means for your problem that one similar sequence might be:
$$\dfrac{1}{6n^2} \quad or \quad  \dfrac{1}{n^2}$$ $\dfrac{1}{n}$ is not good, or better said not helpful. These methods have their root in your epsilon, and their true just because of what Simon said. 
