#### Answer

a) $\lim\limits_{n \to \infty} a_{n}$ = $\lim\limits_{n \to \infty} \frac{2n}{3n+1} = \frac{2}{3}$ ; Convergent .
b) $\lim\limits_{n \to \infty} a_{n} = \frac{2}{3}$ ; the series $\Sigma a_{n} $ diverges.

#### Work Step by Step

Part A requires that we refer to definition in page 736 and take the limit of the sequence. We take n as a common factor from both numerator and denominator and they cancel out each other which leaves us with 2/3 after taking n to infinity.
Part B requires that we use the result from part A and state the test for divergence of series and conclude that since the limit of the sequence gives wasn't zero, the series corresponding to the sequence is divergent.