## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 10: Infinite Sequences and Series - Section 10.5 - Absolute Convergence; The Ratio and Root Tests - Exercises 10.5 - Page 597: 8

Diverges

#### Work Step by Step

Let us consider $a_n=\dfrac{n5^n}{ (2n+3) \ln (n+1)}$ In order to solve the given series we will take the help of Ratio Test. This test states that when the limit $L \lt 1$, the series converges and for $L \gt 1$, the series diverges. $L=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(n+1)5^{n+1}}{ (2(n+1)+3) \ln (n+2)}}{\dfrac{n5^n}{ (2n+3) \ln (n+1)}}|$ $\implies \lim\limits_{n \to \infty}|\dfrac{5(n+1)(2n+3) \ln (n+1)}{n \ln (n+2) (2n+5)}|=[\lim\limits_{n \to \infty}|\dfrac{5(n+1)(2n+3)}{n (2n+5)}|][\lim\limits_{n \to \infty}|\dfrac{ \ln (n+1)}{ \ln (n+2) }|]$ and $L=[5] \times [1]=5 \gt 1$ Thus, the series Diverges by the ratio test.

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