 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/19/c/ii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

For the positive summand

a_{n} = (n!)^{3} e^{3 n} / (3 n)!

, the ratio test gives

\frac{a _{n + 1}}{a _{n}} = \frac{e ^{3} ( n + 1 ) ^{3}}{( 3 n + 1 ) ( 3 n + 2 ) ( 3 n + 3 )} ⟶ \frac{e ^{3}}{27} < 1.

Thus the series converges absolutely.

 New to topics? Read the docs here!