## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises: 15

#### 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.

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