Science:Math Exam Resources/Courses/MATH105/April 2018/Question 04 (c)

MATH105 April 2018

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Question 04 (c)
In this question, if you use the integral test for a series, please explain why the integral test can be used for the series. (c) Find the radius of convergence $R$ of the series: $\sum _{k=0}^{\infty }{\frac {(k+1)}{5^{k}}}(x-9)^{k}$.

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Hint
Consider the ratio test.

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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. We are going to use the ratio test. Since we have $$\lim _{k\to \infty }\left|{\frac {{\frac {(k+2)}{5^{k+1}}}(x-9)^{k+1}}{{\frac {(k+1)}{5^{k}}}(x-9)^{k}}}\right|=\lim _{k\to \infty }{\frac {k+2}{k+1}}\cdot {\frac {5^{k}}{5^{k+1}}}\cdot \left|{\frac {(x-9)^{k+1}}{(x-9)^{k}}}\right|={\frac {1}{5}}\lim _{k\to \infty }{\frac {k+2}{k+1}}|x-9|={\frac {1}{5}}|x-9|,$$ the series $\sum _{k=0}^{\infty }{\frac {(k+1)}{5^{k}}}(x-9)^{k}$ converges if $|x-9|<5$ and diverges if $|x-9|>5$. Therefore, the radius of convergence is $R=5$. Answer: $\color {blue}5$