The ratio test is easier to carry out for this sum. You have
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\dfrac{3^{n+1}(n+1)!}{11\cdot16\cdot21\cdot\cdots\cdot(5n+6)\cdot(5(n+1)+6)}}{\dfrac{3^nn!}{11\cdot16\cdot21\cdot\cdots\cdot(5n+6)}}\right| \\\\\\ = \lim_{n\to\infty}\left|\frac{\dfrac{3(n+1)}{(5(n+1)+6)}}{\dfrac{1}{1}}\right| \\\\\\ = \lim_{n\to\infty}\left|\frac{3n+3}{5n+11}\right| = \frac35 < 1[/tex]
which means the series converges (absolutely).
