Question

Determine whether the sequence with th given \(n\) th term is monotonic. Discuss the boundedness of th sequence. Use a graphing utility to confirm your results. \(a_{n}=\left(-\frac{2}{3}\right)^{n}\)

Short answer

The sequence \(a_{n}=\left(-\frac{2}{3}\right)^{n}\) is not monotonic because it alternates between negative and positive. It is bounded below by 0 and above by \(- \frac{2}{3}\).

Step-by-step solution

Determine Monotonicity

To determine if the sequence \(a_{n}=\left(-\frac{2}{3}\right)^{n}\) is monotonic, compare two consecutive terms, \(a_{n}\) and \(a_{n+1}\), and check if its either strictly increasing or decreasing. \n For \(n=1\) , \(a_{1}=\left(-\frac{2}{3}\right)^{1}= - \frac{2}{3}\) and for \(n=2\), \(a_{2}=\left(-\frac{2}{3}\right)^{2}= \frac{4}{9}\). Since -2/3 is less than 4/9, the sequence is increasing at this point. But to determine overall monotonicity a pattern needs to be established by checking more terms or looking at the general form of the sequence. Considering the power \(n\), if \(n\) is even the term is positive and if \(n\) is odd the term is negative. Therefore for \(n = 3\), \(a_{3}=\left(-\frac{2}{3}\right)^{3}= \left(-\frac{8}{27}\right)\). As expected, \(a_{3}\) is, again, less than \(a_{2}\), so the sequence alternates between negative and positive values. Hence, it is not monotonic.

Boundedness

Next, we need to check the boundedness of the sequence. Boundedness involves checking the maximum and minimum limits of the sequence. Since \(n\) is in the exponent, this sequence will gradually get smaller in magnitude over time regardless if the outcome is positive or negative depending on the parity of \(n\). So, considering \(n\geq1\) (since \(n=0\) would give the result 1 which is not pattern confirming), the sequence is always less than or equal to \(- \frac{2}{3}\) and greater than 0, hence it is bounded below by 0 and above by \(- \frac{2}{3}\).

Confirm with Graphing Utility

This step consists of making a graph of the given sequence. Separate the even \(n\) and odd \(n\), because the sequence is alternatively negative and positive. Plotting the numbers on the x-axis with their corresponding sequence outputs on the y-axis, the behavior of the function can be observed. It should be noted that the sequence values will continue to shrink in magnitude towards 0 as \(n\) increases. This provides a visual confirmation of the boundedness and non-monotonic behavior of this sequence.