Analyzing the Convergence of an Infinite Series
The given series is:
\[ \sum_{n=1}^{\infty} \frac{n!}{(3n)!} \]
We will use the Ratio Test to determine the convergence of the series. The Ratio Test says that for a series \(\sum a_n\), if
\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \]
and
- If \(L < 1\), the series absolutely converges.
- If \(L > 1\) or \(L = \infty\), the series diverges.
- If \(L = 1\), the test is inconclusive.
<p>Let \(a_n = \frac{n!}{(3n)!}\).</p>
<p>We need to compute \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).</p>
<p>\( \left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)!/(3(n+1))!}{n!/(3n)!} = \frac{(n+1)! \cdot (3n)!}{(3n+3)! \cdot n!} = \frac{(n+1)}{(3n+1)(3n+2)(3n+3)} \)</p>
<p>\( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)}{(3n+1)(3n+2)(3n+3)} = \lim_{n \to \infty} \frac{1}{27n^2} = 0 \)</p>
<p>Since \(0 < 1\), by the Ratio Test, the series absolutely converges.</p>
