Question

Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{(n !)^{2}}{(3 n) !} $$

Short answer

The series converges according to the Ratio Test.

Step-by-step solution

Calculate the Next Term in the Series

Find the term \(a_{n+1}\). This is just the term \(a_n\) but with \(n+1\) substituted in where \(n\) was. For our series, \(a_{n+1} = \frac{((n+1) !)^{2}}{(3(n+1)) !}\)

Compute the Ratio of Two Consecutive Terms

Compute the ratio \(a_{n+1}/a_n = \frac{((n+1) !)^{2} / (3(n+1)) !}{(n !)^{2} / (3n) !} = \frac{((n+1)^{2} n!)^{2}}{(3n+3)!} \cdot \frac{(3n)!}{(n!)^{2}} \). This ratio simplifies to \(=\frac{((n+1)^{2})}{(3n+3)(3n+2)(3n+1)}\)

Compute the Limit of the Ratio

Calculate the limit of the ratio as \(n\) approaches infinity, which will give the value the ratio test requires. The limit as \(n\) approaches infinity of \( a_{n+1}/a_{n} \) is \(\lim_{n->\infty} \frac{((n+1)^{2})}{(3n+3)(3n+2)(3n+1)} = 0 \). We find this by recognizing the larger powers of \(n\) in the denominator meaning the limit goes to zero.

Determine Convergence or Divergence

Since the result of this limit is less than 1, by the Ratio Test, the series converges.