Question

Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{n !}\)

Short answer

The sequence \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{n !}\) is divergent.

Step-by-step solution

Express the next term

Express \(a_{n+1}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n + 1)}{(n+1) !}\)

Compute ratio

Compute the ratio \(R= \frac{a_{n+1}}{a_n} ~=~ \frac{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n + 1)}{(n+1) !}}{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{n !}}\). Simplify the ratio to \(R=\frac{2n+1}{n+1}\)

Apply Ratio Test

For the Ratio Test, if the absolute value of \(R\) is less than 1, the series is absolutely convergent. If it is more than 1, it is divergent, and if it is equal to 1, the test is inconclusive. Calculate the limit as \(n\) tends to infinity. The limit here is \( \lim_{n \to \infty} \frac{2n+1}{n+1} = 2 \)

Conclusion

Since the limit of the absolute value of the ratio is greater than 1, it can be concluded that the sequence is divergent