Nice question

The general term of the sequence is $a_n = \dfrac {n^2}{3n^3+500}$ , where $n$ is a positive integer. Treating $a_n$ as continuous and differentiate it with respect to $n$ :

$\begin{aligned} \frac {da_n}{dn} & = \frac {2n(3n^3+500) - n^2(9n^2)}{(3n^3+500)^2} = \frac {1000n-3n^4}{(3n^3+500)^2} \end{aligned}$

Equating $\dfrac {da_n}{dn} = 0$ , we have $n = \sqrt[3]{\dfrac {1000}3}$ and note that $\dfrac {d^2a_n}{dn^2} < 0$ , when $n = \sqrt[3]{\dfrac {1000}3}$ . This means that $a_n$ is maximum when $n = \sqrt[3]{\dfrac {1000}3} \approx 6.934$ and the nearest integer is $n=7$ . Therefore, the largest term is $a_7 = \dfrac {7^2}{3(7^3)+ 500} = \boxed{\dfrac {49}{1529}}$ .