Science:Math Exam Resources/Courses/MATH101/April 2012/Question 05

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MATH101 April 2012

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Question 05
Find, with explanation, the radius of convergence and the interval of convergence of the power series $\sum _{n=0}^{\infty }(-1)^{n}{\frac {(x-1)^{n}}{2^{n}(n+2)}}$

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 figure out for which values of x does the series converge.

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Solution

We will use the ratio test: ${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {\frac {(x-1)^{n+1}}{2^{n+1}(n+3)}}{\frac {(x-1)^{n}}{2^{n}(n+2)}}}\right|\\&=\lim _{n\to \infty }{\frac {|x-1|(n+2)}{2\cdot (n+3)}}\\&=|{\frac {x-1}{2}}|\end{aligned}}$ We want this limit to be $<1$ so that the series converges absolutely. This means $|{\frac {x-1}{2}}|<1$ which means that the radius of convergence is $2$ and that $-1<x<3$. Now we have to test the endpoints. At $x=3$ we get: $\sum _{n=0}^{\infty }{\frac {(-1)^{n}2^{n}}{2^{n}(n+2)}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(n+2)}}$ which converges by alternate series test, if we sum two consecutive terms we get: ${\frac {1}{n}}-{\frac {1}{n+1}}={\frac {1}{n(n+1)}}<{\frac {1}{n^{2}}}$ and the series $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ converges. At $x=-1$ we get $\sum _{n=0}^{\infty }{\frac {(-1)^{n}(-2)^{n}}{2^{n}(n+2)}}=\sum _{n=0}^{\infty }{\frac {1}{(n+2)}}$ which diverges because it is the tail of the harmonic series. So, the interval of convergence is $(-1,3].$

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