Science:Math Exam Resources/Courses/MATH220/April 2005/Question 04

MATH220 April 2005

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Question 04
Define S to be the set of all real numbers x such that exactly one of the two series $\sum _{k=1}^{\infty }\left(x-{\frac {1}{2}}\right)^{k}\quad {\text{ and }}\quad \sum _{k=1}^{\infty }\left(x+{\frac {1}{2}}\right)^{k}$ converges. Write down (with justification) a simple expression for S, using interval notation and set operations such as $\cup$, $\cap$ and / or -.

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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
The series $\sum _{k=1}^{\infty }x^{k}$ converges for all values of x between -1 and 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. The first series $\sum _{k=1}^{\infty }\left(x-{\frac {1}{2}}\right)^{k}$ converges for all values of x such that $\displaystyle -1<x-{\frac {1}{2}}<1$ which we can rewrite as $\displaystyle -1+{\frac {1}{2}}<x<1+{\frac {1}{2}}$ that is $\displaystyle -{\frac {1}{2}}<x<{\frac {3}{2}}$ The second series $\sum _{k=1}^{\infty }\left(x+{\frac {1}{2}}\right)^{k}$ converges for all values of x such that $\displaystyle -1<x+{\frac {1}{2}}<1$ which we can rewrite as $\displaystyle -1-{\frac {1}{2}}<x<1-{\frac {1}{2}}$ that is $\displaystyle -{\frac {3}{2}}<x<{\frac {1}{2}}$ So values of x in the interval (-1/2,3/2) will make the first series converge and values in the interval (-3/2,1/2) will make the second series converge. Hence the values of x which make exactly one of the series convergent cannot be in the intersection of these two intervals. We can now conclude that that the set S is $\displaystyle S=\left(-{\frac {3}{2}},-{\frac {1}{2}}\right]\cup \left[{\frac {1}{2}},{\frac {3}{2}}\right)$

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