Is series Convergent or Divergent? Find sum if convergent.
$\displaystyle \sum\limits_{i=1}^{infty} \frac{n^3}{n(2n+3)^2}$
$\displaystyle \sum\limits_{n=1}^{\infty} \frac{n^3}{n(2n+3)^2}$
$\displaystyle \sum\limits_{n=1}^{\infty} \frac{n^3}{4n^3 + 12n^2+ 9n}$
$\displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{4 + \frac{12}{n}+ \frac{9}{n^2}}$
$\displaystyle \lim_{n\to\infty}\frac{1}{4 + \frac{12}{n}+ \frac{9}{n^2}} = \frac{1}{4} \neq 0 $
so answer is "divergent"?