Science:Math Exam Resources/Courses/MATH101/April 2015/Question 11 (b)

MATH101 April 2015

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Question 11 (b)
Find the interval of convergence for the power series $\sum _{n=1}^{\infty }{\frac {(x-2)^{n}}{n^{\frac {4}{5}}(5^{n}-4)}}.$

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Hint
Recall that to have a convergent series, by ratio test the following must hold: ${\underset {n\rightarrow \infty }{\lim }}{\frac {a_{n+1}}{a_{n}}}=L<1$

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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 apply the ratio test: ${\begin{aligned}{\underset {n\rightarrow \infty }{\lim }}\left\vert {\frac {a_{n+1}}{a_{n}}}\right\vert ={\underset {n\rightarrow \infty }{\lim }}{\dfrac {\left\vert {\frac {(x-2)^{n+1}}{{(n+1)}^{\frac {4}{5}}(5^{n+1}-4)}}\right\vert }{\left\vert {\frac {(x-2)^{n}}{n^{\frac {4}{5}}(5^{n}-4)}}\right\vert }}&={\underset {n\rightarrow \infty }{\lim }}|x-2|{\frac {n^{\frac {4}{5}}(5^{n}-4)}{{(n+1)}^{\frac {4}{5}}(5^{n+1}-4)}}\\&=|x-2|{\underset {n\rightarrow \infty }{\lim }}\left({\frac {n}{n+1}}\right)^{\frac {4}{5}}{\frac {5^{n}(1-{\frac {4}{5^{n}}})}{5^{n+1}(1-{\frac {4}{5^{n+1}}})}}\\\end{aligned}}$ We know ${\underset {n\rightarrow \infty }{\lim }}{\frac {n}{n+1}}=1$ and ${\underset {n\rightarrow \infty }{\lim }}{\frac {4}{5^{n}}}={\underset {n\rightarrow \infty }{\lim }}{\frac {4}{5^{n+1}}}=0$, so we have ${\underset {n\rightarrow \infty }{\lim }}{\frac {5^{n}(1-{\frac {4}{5^{n}}})}{5\cdot 5^{n}(1-{\frac {4}{5^{n+1}}})}}={\frac {1}{5}}$ Therefore; ${\underset {n\rightarrow \infty }{\lim }}\left\vert {\frac {a_{n+1}}{a_{n}}}\right\vert =|x-2|{\frac {1}{5}}=L$. For the series to be convergent we must have: ${\begin{aligned}|x-2|{\frac {1}{5}}<1&\Rightarrow |x-2|<5\\&\Rightarrow -5<x-2<5\\&\Rightarrow -3<x<7\end{aligned}}$. Finally, we need to check for the endpoints: $x=-3,x=7$. We plug each into the series and check the convergence using a suitable test: ${\begin{aligned}x=7\Rightarrow \sum _{n=1}^{\infty }{\frac {(7-2)^{n}}{n^{\frac {4}{5}}(5^{n}-4)}}&=\sum _{n=1}^{\infty }{\frac {5^{n}}{n^{\frac {4}{5}}(5^{n}-4)}}\\&=\sum _{n=1}^{\infty }{\frac {(5^{n}-4)+4}{n^{\frac {4}{5}}(5^{n}-4)}}\\&=\sum _{n=1}^{\infty }{\frac {1}{n^{\frac {4}{5}}}}+{\frac {4}{n^{\frac {4}{5}}(5^{n}-4)}}\\&=\sum _{n=1}^{\infty }{\frac {1}{n^{\frac {4}{5}}}}+\sum _{n=1}^{\infty }{\frac {4}{n^{\frac {4}{5}}(5^{n}-4)}}\end{aligned}}$ The first series diverges by Comparison Test with $\sum _{n=1}^{\infty }{\frac {1}{n}}$ because ${\frac {1}{n^{\frac {4}{5}}}}>{\frac {1}{n}}$, so $x=7$ is NOT included in the interval of convergence. Similarly: ${\begin{aligned}x=-3\Rightarrow \sum _{n=1}^{\infty }{\frac {(-3-2)^{n}}{n^{\frac {4}{5}}(5^{n}-4)}}&=\sum _{n=1}^{\infty }{\frac {(-1)^{n}\cdot 5^{n}}{n^{\frac {4}{5}}(5^{n}-4)}}\\&=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n^{\frac {4}{5}}}}+{\frac {4(-1)^{n}}{n^{\frac {4}{5}}(5^{n}-4)}}\\&=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n^{\frac {4}{5}}}}+\sum _{n=1}^{\infty }{\frac {4(-1)^{n}}{n^{\frac {4}{5}}(5^{n}-4)}}\end{aligned}}$ This time we need to use the Alternating Series Test: It is easy to see that both of the summands in the two series i.e. ${\frac {1}{n^{\frac {4}{5}}}}$, and ${\frac {4}{n^{\frac {4}{5}}(5^{n}-4)}}$ converge to zero as $n\rightarrow \infty$, and they are both decreasing, so from the alternating series test we conclude that they are convergent, which means that $x=-3$ is included in the interval of convergence. Interval of convergence: ${\color {blue}[-3,7)}$