## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 27

#### Answer

The series is divergent.

#### Work Step by Step

If we look at the terms of the infinite sum we can see a pattern of an increase of 3 in the denominator with each progression in the terms, so first term is $\frac{1}{3}$ and second is$\frac{1}{6}$ and so on. This means that there is some sort of relation between the index of terms and the increase of 3 in the denominator of each term. Our given series can be written explicitly as the infinite series $\Sigma \frac{1}{3n}$ = $\frac{1}{3} \Sigma \frac{1}{n}$ (This is by thoerem 8 in page 754 which states that we can take the constant out of the sigma notation). After our last step, we can see that the series we have is a harmonic series multiplied by $\frac{1}{3}$ which is known to be divergent (p-series with p=1, check page 762 later in the chapter), and this means that our original series is divergent.

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