Proof using convergence of a sequence I have a simple question just to see if my idea is right.
Suppose we have a sequence $(a_n)=\left(\frac{1}{6n^2+1}\right)$.  I want to show that the $\lim (a_n) = 0$.  Using the definition of convergence of a sequence, I just have to show that $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that when $n > N, |\frac{1}{6n^2+1}| < \epsilon$ 
My question is, since $|\frac{1}n| > |1/(6n^2+1)|$, can I just show $|\frac{1}n|<\epsilon$?  I'm pretty sure I can, but I just wanted to check my logic here.  I can do the proof the rest of the way regardless.
 A: You're right!  
Once you show that $\left |\frac{1}{n} \right | < \epsilon$, then since $\left |\frac{1}{6n^{2} + 1} \right | < \left |\frac{1}{n}\right |$, it follows that $\left |\frac{1}{6n^{2} + 1} \right | < \epsilon$.
So a basic proof would look like: 

Let $\epsilon > 0$.  We know we can find $N$ such that for all $n \geq N$, $\left | \frac{1}{n} \right | < \epsilon$.  But for all $n$, we have $\left |\frac{1}{6n^{2} + 1} \right | < \left |\frac{1}{n}\right |$.  Thus, for all $n \geq N$, we get $\left |\frac{1}{6n^{2} + 1} \right | < \left |\frac{1}{n}\right | < \epsilon$.  So, we have shown that for every $\epsilon > 0$, $\exists N$ such that $n > N$ implies $\left |\frac{1}{6n^{2} + 1} \right | < \epsilon$, and this is precisely what it means for $\left ( \frac{1}{6n^{2} + 1} \right )$ to converge to $0$.

A: To expand upon David's answer, consider the fact that $1/n$ is larger than $1/(6n^2+1)$ for all $n\geq N$, where $N$ is some integer (in this case, you can verify $N=1$). Both sequences are nonnegative, so now you can apply the Squeeze theorem.
