find the limit of the following sequence or determine that the sequence diverges. {(-1)^n / (2n + 9)} select…

Explanation:

Step1: Recall the squeeze - theorem

We know that (-1\leqslant(- 1)^n\leqslant1) for all (n\in N). Then (\frac{-1}{2n + 9}\leqslant\frac{(-1)^n}{2n+9}\leqslant\frac{1}{2n + 9}).

Step2: Find the limits of the bounding sequences

Calculate (\lim_{n\rightarrow\infty}\frac{-1}{2n + 9}). Let (t = 2n+9), as (n\rightarrow\infty), (t\rightarrow\infty). Then (\lim_{n\rightarrow\infty}\frac{-1}{2n + 9}=\lim_{t\rightarrow\infty}\frac{-1}{t}=0). Also, calculate (\lim_{n\rightarrow\infty}\frac{1}{2n + 9}). Let (t = 2n + 9), as (n\rightarrow\infty), (t\rightarrow\infty). Then (\lim_{n\rightarrow\infty}\frac{1}{2n + 9}=\lim_{t\rightarrow\infty}\frac{1}{t}=0).

Step3: Apply the squeeze - theorem

Since (\lim_{n\rightarrow\infty}\frac{-1}{2n + 9}=0) and (\lim_{n\rightarrow\infty}\frac{1}{2n + 9}=0), by the squeeze - theorem (\lim_{n\rightarrow\infty}\frac{(-1)^n}{2n+9}=0).

Answer:

A. The limit of the sequence is (0)