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

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MATH101 April 2015

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

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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

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Solution

We apply the ratio test: $\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \mathrm{\infty }}{\lim }\frac{\left|\frac{(x-2{)}^{n+1}}{{(n+1)}^{\frac{4}{5}}(5^{n+1}-4)}\right|}{\left|\frac{(x-2{)}^n}{n^{\frac{4}{5}}(5^n-4)}\right|}& =\underset{n\to \mathrm{\infty }}{\lim }|x-2|\frac{n^{\frac{4}{5}}(5^n-4)}{{(n+1)}^{\frac{4}{5}}(5^{n+1}-4)}\\ & =|x-2|\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{n}{n+1}\right)}^{\frac{4}{5}}\frac{5^n(1-\frac{4}{5^n})}{5^{n+1}(1-\frac{4}{5^{n+1}})}\\ \end{array}$ We know $\underset{n\to \mathrm{\infty }}{\lim }\frac{n}{n+1}=1$ and $\underset{n\to \mathrm{\infty }}{\lim }\frac{4}{5^n}=\underset{n\to \mathrm{\infty }}{\lim }\frac{4}{5^{n+1}}=0$, so we have $\underset{n\to \mathrm{\infty }}{\lim }\frac{5^n(1-\frac{4}{5^n})}{5\cdot 5^n(1-\frac{4}{5^{n+1}})}=\frac{1}{5}$ Therefore; $\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{a_{n+1}}{a_n}\right|=|x-2|\frac{1}{5}=L$. For the series to be convergent we must have: $\begin{array}{rl}|x-2|\frac{1}{5}<1& \Rightarrow |x-2|<5\\ & \Rightarrow -5<x-2<5\\ & \Rightarrow -3<x<7\end{array}$. 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: $\begin{array}{rl}x=7\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{(7-2{)}^n}{n^{\frac{4}{5}}(5^n-4)}& =\sum _{n=1}^{\mathrm{\infty }}\frac{5^n}{n^{\frac{4}{5}}(5^n-4)}\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{(5^n-4)+4}{n^{\frac{4}{5}}(5^n-4)}\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{\frac{4}{5}}}+\frac{4}{n^{\frac{4}{5}}(5^n-4)}\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{\frac{4}{5}}}+\sum _{n=1}^{\mathrm{\infty }}\frac{4}{n^{\frac{4}{5}}(5^n-4)}\end{array}$ The first series diverges by Comparison Test with $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ because $\frac{1}{n^{\frac{4}{5}}}>\frac{1}{n}$, so $x=7$ is NOT included in the interval of convergence. Similarly: $\begin{array}{rl}x=-3\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{(-3-2{)}^n}{n^{\frac{4}{5}}(5^n-4)}& =\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n\cdot 5^n}{n^{\frac{4}{5}}(5^n-4)}\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^{\frac{4}{5}}}+\frac{4(-1{)}^n}{n^{\frac{4}{5}}(5^n-4)}\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^{\frac{4}{5}}}+\sum _{n=1}^{\mathrm{\infty }}\frac{4(-1{)}^n}{n^{\frac{4}{5}}(5^n-4)}\end{array}$ 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 $n\to \mathrm{\infty }$, 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: $[-3,7)$

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