Science:Math Exam Resources/Courses/MATH105/April 2017/Question 05 (e)

MATH105 April 2017

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Question 05 (e)
Find the radius of convergence of the power series $\sum _{k=1}^{\infty }{\frac {(x-1)^{2k}}{k(5^{k})}}$. .

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Hint
In order to determine the radius of convergence of a power series, it is typically correct to use 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 will use the ratio test to determine the radius of convergence. The ratio of the $k+1$ term to the $k$ term is ${\frac {\frac {(x-1)^{2(k+1)}}{(k+1)5^{k+1}}}{\frac {(x-1)^{2k}}{k5^{k}}}}$. We will invert and multiply: ${\frac {(x-1)^{2k+2}k5^{k}}{(x-1)^{2k}(k+1)5^{k+1}}}$. We cancel out a factor of $(x-1)^{2k}$ and $5^{k}$. This gives us ${\frac {(x-1)^{2}k}{5(k+1)}}.$ Now, we take the limit as $k\to \infty$. $\lim _{k\to \infty }{\frac {(x-1)^{2}}{5}}{\frac {k}{k+1}}={\frac {(x-1)^{2}}{5}}.$ This will be less than 1 in absolute value whenever the absolute value of the numerator is less than the absolute value of the denominator. This gives us the inequality $|(x-1)^{2}|<5$, which happens whenever $|x-1|<{\sqrt {5}}.$ Therefore, the radius of convergence is ${\sqrt {5}}.$ Answer: $\color {blue}{\sqrt {5}}$