Question

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

Short answer

The sequence diverges.

Step-by-step solution

Identify the nth term and the (n+1)th term

The nth term provided in the exercise is \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{(2 n)^{n}}\). To find the (n+1)th term, simply replace 'n' with 'n+1' to get \(a_{n+1}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n+1-1)}{(2 (n+1))^{n+1}} = \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot 2n \cdot 2n+1}{(2n+2)^{n+1}}\).

Apply the Ratio Test

Using the Ratio Test to determine the convergence or divergence of the sequence involves dividing the (n+1)th term by the nth term, and taking the limit as n tends to infinity. \(\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}} =\lim_{n\to\infty} \frac{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot 2n \cdot 2n+1}{(2n+2)^{n+1}}}{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{(2 n)^{n}}} = \lim_{n\to\infty} \frac{(2n+1)(2n)^n}{(2n+2)^{n+1}}\).

Simplify the expression

Simplify the expression obtained in Step 2 to render its limit calculation easier. This gives \(\lim_{n\to\infty} \frac{(2n+1)}{(2n+2)}\).

Calculate the limit

Calculate the limit as n tends to infinity for \(\lim_{n\to\infty} \frac{(2n+1)}{(2n+2)}\). As both numerator and denominator tend to infinity, this is an indeterminate form of type \(\frac{\infty}{\infty}\). Dividing both numerator and denominator by n yields \(\lim_{n\to\infty} \frac{2+(1/n)}{2+(2/n)}\). Taking the limit as n tends to infinity gives the ratio \(\frac{2}{2} = 1\).

Determine convergence or divergence

As the value of the limit (ratio) is 1, which is not less than 1, so according to the Ratio Test, the sequence diverges.