Science:Math Exam Resources/Courses/MATH101/April 2011/Question 08/Solution 1

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 $[1,3)$ .