Science:Math Exam Resources/Courses/MATH101/April 2011/Question 08

MATH101 April 2011

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Question 08
Find (a) the radius of convergence, and (b) the interval of convergence of the following power series, carefully justifying your answer: $\sum _{n=0}^{\infty }{\frac {(x-2)^{n}}{\ln(n+2)}}$

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Hint
a.) Ratio Test b.) Check end points

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To find the radius of convergence, we use the ratio test. Let $a_{n}={\frac {\ln(n+2)}{|x-2|^{n}}}.$ ${\begin{aligned}\left|{\frac {a_{n+1}}{a_{n}}}\right|&={\frac {|x-2|^{n+1}}{\ln(n+1+2)}}\cdot {\frac {\ln(n+2)}{|x-2|^{n}}}\\&=|x-2|{\frac {\ln(n+2)}{\ln(n+3)}}\to |x-2|{\text{ as }}n\to \infty .\\\end{aligned}}$ To see that ${\frac {\ln(n+2)}{\ln(n+3)}}\to 1{\text{ as }}n\to \infty ,$ note that ${\begin{aligned}\left|{\frac {\ln(n+2)}{\ln(n+3)}}-1\right|&=\left|{\frac {\ln(n+2)-\ln(n+3)}{\ln(n+3)}}\right|=\left|{\frac {\ln \left({\frac {n+2}{n+3}}\right)}{\ln(n+3)}}\right|\end{aligned}}$ Since logarithm is continuous on its domain, the numerator satisfies ${\begin{aligned}\lim _{n\to \infty }\ln \left({\frac {n+2}{n+3}}\right)&=\ln \left(\lim _{n\to \infty }{\frac {n+2}{n+3}}\right)=\ln \left(\lim _{n\to \infty }{\frac {1+2/n}{1+3/n}}\right)=\ln(1)=0.\end{aligned}}$ In the denominator, $\ln(n+3)\to \infty$ as $n\to \infty$. We conclude that $\left|{\frac {\ln(n+2)}{\ln(n+3)}}-1\right|\to 0$; that is, ${\frac {\ln(n+2)}{\ln(n+3)}}\to 1,$ as $n\to \infty .$ By the ratio test, it follows that the series is absolutely convergent if and only if $|x-2|<1.$ Therefore, the radius of convergence is equal to 1. In order to determine the interval of convergence, it remains to determine whether the series is convergent at the endpoints, $x=2\pm 1.$ At $x=3,$ the series is equal to $\sum _{n=1}^{\infty }{\frac {(3-2)^{n}}{\ln(n+2)}}=\sum _{n=1}^{\infty }{\frac {1}{\ln(n+2)}}.$ By comparison with $\sum _{n=1}^{\infty }{\frac {1}{n}},$ this series is divergent. (To compare, note that ${\frac {1}{\ln(n+2)}}>{\frac {1}{n+2}}\geq {\frac {1}{3n}}.$) At $x=1$ the series is equal to $\sum _{n=1}^{\infty }{\frac {(1-2)^{n}}{\ln(n+2)}}=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\ln(n+2)}}$ By the alternating series test, this converges. Hence, the interval of convergence is $[1,3)$ .

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