Question

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. \(a_{n}=\frac{3 n}{n+2}\)

Short answer

The sequence \(a_{n}=\frac{3 n}{n+2}\) is an increasing (thus monotonic) and bounded sequence. The lower bound of the sequence is 0 and the upper bound is 3.

Step-by-step solution

Determine the Monotonicity

For a sequence to be monotonic, it must be entirely non-increasing or non-decreasing. Their difference \(a_{n+1} - a_n\) can indicate whether the sequence is increasing or decreasing. If \(a_{n+1} - a_n > 0\) then the sequence is increasing. If \(a_{n+1} - a_n < 0\), then the sequence is decreasing.

Computed the Difference \(a_{n+1} - a_{n}\)

Substitute \(n+1\) for \(n\) in the given sequence \(a_{n}\) to get \(a_{n+1}=\frac{3(n+1)}{(n+1)+2}=\frac{3n+3}{n+3}\), then find the difference. \(a_{n+1} - a_{n} =\frac{3n+3}{n+3} - \frac{3n}{n+2}\). Solve this difference, the result is: \(a_{n+1} - a_{n} = \frac{6}{(n+2)(n+3)}\). Since \(n >= 1\), the difference is always positive, hence the sequence is increasing.

Determine the Boundedness of the Sequence

The sequence is bounded if there are upper or lower limits. It is clear that the sequence is positive for all values of \(n >= 1\), hence it is lower bounded by 0. Next, if you take the limit of the sequence as \(n\) approaches infinity, i.e. \(\lim_{n \to \infty} \frac{3n}{n+2} =3\), it is clear that the sequence is also upper bounded by 3.

Confirm the Results with a Graphing Utility

Plotting the sequence with a graphing utility will help confirm the results. The x-axis can be considered as 'n' (the term number) and the y-axis represents the sequence \(a_{n}\). The plot should indicate the sequence as increasing and bounded by the lines \(y=0\) (lower bound) and \(y=3\) (upper bound).