## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 9 - Infinite Series - 9.2 Monotone Sequences - Exercises Set 9.2 - Page 613: 7

#### Answer

strictly increasing

#### Work Step by Step

Given $$\left\{ \frac{n}{2n+1}\right\}_{1}^{\infty}$$ Since $a_n=\dfrac{n}{2n+1},\ \ a_{n+1}= \dfrac{n+1}{2n+3}$, then \begin{align*} \frac{a_{n+1}}{a_n}&=\dfrac{n+1}{2n+3}\frac{2n+1}{n}\\ &= \dfrac{2n^2+3n+1}{2n^2+3n}\\ &=1+\dfrac{ 1}{2n^2+3n}>1 \end{align*} so, the sequence is eventually strictly increasing

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