Solution
= Solution
For the positive summand $a_n=(n!)^3e^{3n}/(3n)!$, the <ratio test> gives
$$
\frac{a_{n+1}}{a_n}=\frac{e^3(n+1)^3}{(3n+1)(3n+2)(3n+3)}\longrightarrow\frac{e^3}{27}<1.
$$
Thus the series \b[converges absolutely].
= Solution
For the positive summand $a_n=(n!)^3e^{3n}/(3n)!$, the <ratio test> gives
$$
\frac{a_{n+1}}{a_n}=\frac{e^3(n+1)^3}{(3n+1)(3n+2)(3n+3)}\longrightarrow\frac{e^3}{27}<1.
$$
Thus the series \b[converges absolutely].