Fluctuating series!

$The\quad general\quad term\quad can\quad be\quad written\quad as\\ \\ \quad \quad \quad \quad \quad \quad \quad { T }_{ n }\quad =\quad \frac { { n }^{ 2 } }{ { 500+\quad 3n }^{ 3 } } \\ Differentiating\quad it\quad with\quad respect\quad to\quad 'n'\\ and\quad putting\quad derivative\quad =\quad 0\quad we\quad get\quad \\ maxima\quad at\quad n=7\\ \\ Since\quad { T }_{ 7 }\quad =\quad \frac { 49 }{ 1529 } \quad we\quad get\\ a+b=1578$

The general term is $a_n = \dfrac {n^2}{3n^3+500}$ . Since $a_n > 0$ , maximum $a_n$ occurs when $\dfrac 1{a_n}$ is minimum. And the minimum occurs when

$\begin{aligned} \frac d{dn} \left(\frac 1{a_n}\right) & = 0 \\ \frac d{dn} \left(\frac {3n^3+500}{n^2} \right) & = 0 \\ \frac d{dn} \left(3n+\frac {500}{n^2} \right) & = 0 \\ 3 - \frac {1000}{n^3} & = 0 \\ 3n^3 & = 1000 \\ \implies n & \approx 6.934 = \color{#3D99F6}{7 \quad \small \text{the nearest integer.}} \end{aligned}$

We check that $\dfrac {d^2}{dn^2} \left(\dfrac 1{a_n}\right) \bigg|_{n=\sqrt[3]{\frac {1000}3}} > 0$ $\implies \dfrac 1{a_7}$ is minimum and hence $a_7 = \dfrac {7^2}{3(7^3)+500} = \dfrac {49}{1529}$ is maximum.

$\implies a + b = 49+1529 = \boxed{1578}$

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Note that it is easier to evaluate $\frac d{dn} \left(\frac 1{a_n}\right)$ than $\frac d{dn} a_n$ .