Question - Analyzing the Convergence of an Infinite Series
Solution:
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.
Let \(a_n = \frac{n!}{(3n)!}\).
We need to compute \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
\( \left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)!/(3(n+1))!}{n!/(3n)!} = \frac{(n+1)! \cdot (3n)!}{(3n+3)! \cdot n!} = \frac{(n+1)}{(3n+1)(3n+2)(3n+3)} \)
\( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)}{(3n+1)(3n+2)(3n+3)} = \lim_{n \to \infty} \frac{1}{27n^2} = 0 \)
Since \(0 < 1\), by the Ratio Test, the series absolutely converges.
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