# Using the definition of convergence, how do you prove that the sequence lim 1/(6n^2+1)=0 converges?

Jun 18, 2018

Given any number $\epsilon > 0$ choose $M > \frac{1}{\sqrt{6 \epsilon}}$, with $M \in \mathbb{N}$.

Then, for $n \ge M$ we have:

$6 {n}^{2} + 1 > 6 {n}^{2} > 6 {M}^{2} \ge \frac{6}{6 \epsilon} = \frac{1}{\epsilon}$

and so:

$n \ge M \implies \frac{1}{6 {n}^{2} + 1} < \epsilon$

which proves the limit.