Science:Math Exam Resources/Courses/MATH101/April 2017/Question 06 (b)

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MATH101 April 2017

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Question 06 (b)
For which values of x does the series $\sum _{n=1}^{\infty }{\frac {2}{3^{n}(x-1)^{n+1}}}$ converge?

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?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use the ratio test to determine the intervals where this series converge

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Solution
Apply the ratio test. Since we have $a_{n}={\frac {2}{3^{n}(x-1)^{n+1}}}.$ and ${\frac {a_{n+1}}{a_{n}}}={\frac {2}{3^{n+1}(x-1)^{n+2}}}\cdot {\frac {3^{n}(x-1)^{n+1}}{2}}={\frac {1}{3(x-1)}},$ the following three situations are given $\lim _{n\rightarrow \infty }\left\|{\frac {a_{n+1}}{a_{n}}}\right\|{\begin{cases}>1,&{\text{the series diverges}}\\=1,&{\text{inconclusive}}\\<1,&{\text{the series converges}}\\\end{cases}}$ Therefore, for ${\textstyle \|x-1\|>{\frac {1}{3}}}$ , we have ${\textstyle \lim _{n\rightarrow \infty }\|{\frac {a_{n+1}}{a_{n}}}\|={\frac {1}{3\|x-1\|}}<1}$ , so that the series converges. Now, we determine the convergence of series when $\|x-1\|={\frac {1}{3}}$ . When $x-1={\frac {1}{3}}$ and $x-1=-{\frac {1}{3}}$ , we have $a_{n}={\frac {2}{3^{n}(x-1)^{n+1}}}={\frac {2}{3^{n}\left({\frac {1}{3}}\right)^{n+1}}}=6$ and $a_{n}={\frac {2}{3^{n}(x-1)^{n+1}}}={\frac {2}{3^{n}\left(-{\frac {1}{3}}\right)^{n+1}}}=(-1)^{n+1}6$ , respectively. Since in both case, $\lim _{n\to \infty }a_{n}\neq 0$ , by the divergence test, the series doesn't converges. To summarize, the series converges on ${\textstyle \color {blue}\{x:\|x-1\|>{\frac {1}{3}}\}=\{x<{\frac {2}{3}}{\text{ or }}x>{\frac {4}{3}}\}.}$