To do this, we need to find the limit as $n$ approaches infinity of $a_n$.
\begin{align*}
\lim_{n\to\infty} a_n &= \lim_{n\to\infty} \frac{2n}{3n+1} \\
&= \lim_{n\to\infty} \frac{2}{3+\frac{1}{n}} \quad \text{(dividing numerator and denominator by $n$)} \\
&=
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