Science:Math Exam Resources/Courses/MATH103/April 2017/Question 04

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MATH103 April 2017

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Question 04
Find all $x$ such that the series $\sum _{n=1}^{\infty }{\frac {(n-1)^{2}(x^{2}-1)^{n}}{3^{n}n}}$ converges.

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
Figure out the convergence tests that best suit this problem. To do this, note that exponential functions grow much faster than any polynomial.

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Solution
Let $\sum _{n=1}^{\infty }{\frac {(n-1)^{2}(x^{2}-1)^{n}}{3^{n}n}}=\sum _{n=1}^{\infty }a_{n}$ Use the Ratio Test. This requires us to calculate $L=\lim _{n\to \infty }\|{\frac {a_{n+1}}{a_{n}}}\|$ . Doing so, we obtain, $L=\lim _{n\to \infty }\|{\dfrac {{\frac {n^{2}}{n+1}}({\frac {x^{2}-1}{3}})^{n+1}}{{\frac {(n-1)^{2}}{n}}({\frac {x^{2}-1}{3}})^{n}}}\|=\lim _{n\to \infty }{\frac {n^{2}}{(n-1)^{2}}}{\frac {n}{n+1}}\|{\frac {x^{2}-1}{3}}\|=\|{\frac {x^{2}-1}{3}}\|$ One can check that $\|{\frac {x^{2}-1}{3}}\|<1{\text{ if }}\|x\|<2$ $\|{\frac {x^{2}-1}{3}}\|=1{\text{ if }}\|x\|=2$ $\|{\frac {x^{2}-1}{3}}\|>1{\text{ if }}\|x\|>2$ So by the Ratio Test, the sum converges if $\|x\|<2$ and the sum diverges if $\|x\|>2$ . When $\|x\|=2$ (i.e., $x=2$ or $x=-2$ ), the sum becomes $\sum _{n=1}^{\infty }{\frac {(n-1)^{2}}{n}}$ . Since $\lim _{n\to \infty }\|a_{n}\|=\lim _{n\to \infty }{\frac {(n-1)^{2}}{n}}=\lim _{n\to \infty }{\frac {n-1}{n}}(n-1)=\infty$ , by the Term Test for divergence, the series diverges when $\|x\|=2$ . Thus, the answer is as follows. $\color {blue}{\text{The series converges when }}-2<x<2.$