Question

In Exercises \(63-72,\) determine whether the sequence with the given \(n\) th term is monotonic. Discuss the boundedness of the sequence. Use a graphing utility to confirm your results. \(a_{n}=\frac{n}{2^{n+2}}\)

Short answer

The sequence \(a_{n}=\frac{n}{2^{n+2}}\) is monotonic and bounded. It is non-increasing or monotonic since the difference between any two consecutive terms consistently decreases. It is bounded by 0 and \(\frac{1}{8}\) as every term of the sequence is less than or equal to \(\frac{1}{8}\).

Step-by-step solution

Check if the Sequence is Monotonic

A sequence is monotonic if it's either entirely non-increasing or non-decreasing. To check this we'll examine the difference between a term at position (n+1) and a term at position n. If the value is consistently decreasing, then the sequence is non-increasing, if it's consistently increasing then the sequence is non-decreasing. So we need to calculate \( \frac{n+1}{2^{n+3}} - \frac{n}{2^{n+2}} = \frac{2n + 2 - 2n}{2^{n+3}} = \frac{2}{2^{n+3}}\). Since n can be any positive number, the value will consistently decrease, showing the sequence is non-increasing or monotonic.

Check if the Sequence is Bounded

A sequence is bounded if there is a real number M such that every term of the sequence is less than or equal to M. Since the numerator of the sequence grows linearly and the denominator grows exponentially, the sequence values will get smaller each step. Therefore, the sequence is bounded by 0 on the left (smaller side), and the starting term \(\frac{1}{8}\) on the right (larger side). So we can certainly say that every term of the sequence will be less than or equal to \(\frac{1}{8}\), thus it is bounded.

Use a Graphing Utility for Visual Confirmation

Plotting the sequence using a graphing utility should show that the sequence is decreasing (from examination of the graph slope) and gets closer to zero with increasing n (indicating that the sequence is bounded). Important to note that physical plotting isn't necessary for solving this problem but a useful skill to interpret and validate the results.