Question

Limits of sequences: Determine whether the sequences that follow are bounded, monotonic and/or eventually monotonic.

Determine whether each sequence converges or diverges. If the sequence converges, find its limit.

$\left\{\frac{k!}{1·3·5···(2k-1)}\right\}$

Short answer

• The sequence is montonically decresing and bounded.
• The limit of the sequence is 0.

Step-by-step solution

Step 1. Given Information

The given sequence is $\left\{\frac{k!}{1·3·5···(2k-1)}\right\}$.

Step 2. Apply Ratio Test

• Find the ratio $\frac{a_{k+1}}{a_k}$.

$\begin{array}{rcl}\frac{\frac{(k+1)!}{1·3·5···(2k-1)(2k+1)}}{\frac{k!}{1·3·5···(2k-1)}}& =& \frac{(k+1)!}{1·3·5···(2k-1)(2k+1)}\times \frac{1·3·5···(2k-1)}{k!}\\ & =& \frac{k+1}{2k+1}\end{array}$

• As $\frac{k+1}{2k+1}<1$ for all k, the sequence is strictly monotonically decreasing.

Step 3. Check for Boundedness and Limit

• All the terms of the sequence are positive and the sequence is decreasing, it is bounded below by zero, and because the sequence is strictly decreasing, the first term, 1 will be the upper bound for the sequence.
• Therefore, the sequence is bounded.

• The limit of the sequence will be 0, that is the lower bound of the decreasing sequence,