Consider the sequence \displaystyle{\lbrace a_n \rbrace = \left\lbrace {\frac{\left(-1\right)^{n}\cdot 9n}{n+1}} \right\rbrace } . Graph this sequence and use your graph to help you answer the following questions.

- Is the sequence
\lbrace a_n \rbrace bounded above by a function? If it is, enter the function of the variablen that provides the “best” and “most obvious” upper bound; otherwise, enter DNE for does not exist. - What is the limit of the function from part (a) as
n \to \infty ?Enter a number, or enter DNE. - Is the sequence
\lbrace a_n \rbrace bounded below by a function? If it is, enter the function of the variablen that provides the “best” and “most obvious” lower bound; otherwise, enter DNE for does not exist. - What is the limit of the function from part (c) as
n \to \infty ?Enter a number, or enter DNE. - Is the sequence
\lbrace a_n \rbrace bounded above by a number?Enter a number or enter DNE. - Is the sequence
\lbrace a_n \rbrace bounded below by a number?Enter a number or enter DNE. - Select all that apply: The sequence
\lbrace a_n \rbrace is**A.**bounded below.**B.**unbounded.**C.**bounded above.**D.**bounded.

The sequence \lbrace a_n \rbrace is

** A. ** decreasing.

** B. ** alternating

** C. ** increasing.

** D. ** none of the above

- The sequence
\lbrace a_n \rbrace ischoose convergent divergent - The limit of the sequence
\lbrace a_n \rbrace is. Enter a number or DNE.

- When you first look at the sequence
\displaystyle \left\lbrace \frac{\left(-1\right)^{n}\cdot 9n}{n+1} \right\rbrace , you expect it to**A.**diverge because alternating sequences always diverge.**B.**converge to both-9 and9 because the odd index terms tend to-9 asn \to \infty , while the even index terms tend to9 asn \to \infty .**C.**diverge because the odd index terms tend to-9 asn \to \infty , while the even index terms tend to9 asn \to \infty , so there is not one single value for the limit of the sequence.

- When you first look at the sequence
\displaystyle \left\lbrace \frac{\left(-1\right)^{n}\cdot 9}{n+1} \right\rbrace , you expect it to**A.**diverge because the odd index terms tend to-9 asn \to \infty , while the even index terms tend to9 asn \to \infty , so there is not one single value for the limit of the sequence.**B.**converge to0 because\frac{9}{n+1} \leq \frac{\left(-1\right)^{n}\cdot 9}{n+1} \leq \frac{9}{n+1} and both\frac{-9}{n+1} and\frac{9}{n+1} converge to0 asn \to \infty .**C.**diverge because alternating sequences always diverge.

- If a sequence is alternating, it
choose must may or may not cannot