Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises - Page 766: 22

Answer

$\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ is convergent.

Work Step by Step

Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{\ln n}{n^2}\] \[\Rightarrow a_n=\frac{\ln n}{n^2}\] $a_n$ is positive term and monotonically decreasing. So Integral test is applicable. Let \[I=\int_{1}^{\infty}\frac{\ln x}{x^2}dx\] \[I=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{\ln x}{x^2}dx\;\;\;\;\;\;\;\;\;\;\ldots(1)\] Let \[I_1=\int\frac{\ln x}{x^2}dx\] Substitute $y=\ln x\;\Rightarrow x=e^y$ \[\Rightarrow dy=\frac{1}{x}dx\] \[\Rightarrow I_1=\int\frac{y}{e^y}dy=\int ye^{-y}dy\] Using integration by parts \[\Rightarrow I_1=-ye^{-y}+\int e^{-y}dy\] \[\Rightarrow I_1=-ye^{-y}-e^{-y}=-(y+1)e^{-y}\] \[\Rightarrow I_1=-(\ln x+1)e^{-\ln x}=-\frac{\ln x+1}{x}\;\;\;\;\;\;\;\;\;\;\ldots (2)\] Using (2) in (1) \[I=\lim_{t\rightarrow\infty}\left[-\frac{\ln x+1}{x}\right]_{1}^{t}\] \[\Rightarrow I=\lim_{t\rightarrow\infty}\left[-\frac{\ln t+1}{t}+\frac{\ln 1+1}{1}\right]\] \[\Rightarrow I=\lim_{t\rightarrow\infty}\left[-\frac{\ln t+1}{t}+1\right]\] \[\Rightarrow I=\lim_{t\rightarrow\infty}\left[-\frac{\ln t+1}{t}\right]+1\] Which is $\displaystyle\frac{\infty}{\infty}$ case, apply L' Hopitals rule \[\Rightarrow I=\lim_{t\rightarrow\infty}\left[-\frac{\frac{1}{t}}{1}\right]+1\] \[\Rightarrow I=0+1=1\] Which is finite so $I$ is convergent. By integral test, $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ is convergent.
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