Science:Math Exam Resources/Courses/MATH105/April 2013/Question 06 (a)

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MATH105 April 2013

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Question 06 (a)
Find the interval of convergence of the following series: $\sum _{k=1}^{\infty }{\frac {(x+1)^{2k}}{k^{2}9^{k}}}$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Try applying a series convergence test to determine which values of $x$ will make the series converge. Which series test works best in this situation (that is, a series with a variable x?)

Hint 2
Use the ratio test.

Hint 3
Don't forget to check endpoints!

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

To determine the interval of convergence, we apply the ratio test. Let $a_{k}(x)={\frac {(x+1)^{2k}}{k^{2}9^{k}}}.$ To determine the interval of convergence, we must find the values of $x$ such that the value of the limit below is less than 1 (and then we need to check the endpoints). Evaluating gives ${\begin{aligned}\lim _{k\rightarrow \infty }\left|{\frac {a_{k+1}(x)}{a_{k}(x)}}\right|&=\lim _{k\rightarrow \infty }\left|{\frac {(x+1)^{2k+2}}{(k+1)^{2}9^{k+1}}}\div {\frac {(x+1)^{2k}}{k^{2}9^{k}}}\right|\\&=\lim _{k\rightarrow \infty }\left|{\frac {(x+1)^{2k+2}}{(k+1)^{2}9^{k+1}}}{\frac {k^{2}9^{k}}{(x+1)^{2k}}}\right|\\&=\lim _{k\rightarrow \infty }\left|{\frac {1}{9}}\left({\frac {k}{k+1}}\right)^{2}(x+1)^{2}\right|\\&=\left|{\frac {(x+1)^{2}}{9}}\right|\\&={\frac {(x+1)^{2}}{9}}\end{aligned}}$ where we drop the absolute value signs, since $[(x+1)^{2}]/9$ is always positive. Now, when the above limit is less than one, we have: ${\begin{aligned}{\frac {(x+1)^{2}}{9}}&<1\\(x+1)^{2}&<9\\-3<x+1&<3\\-4<x&<2.\end{aligned}}$ Similarly, we know what when ${\frac {(x+1)^{2}}{9}}>1$ that is, when $x<-4$ and when $x>2$, we have that the series diverges. So all that is left is to check the two endpoints. When $x=-4$, we have ${\begin{aligned}\sum _{k=1}^{\infty }a_{k}(-4)&=\sum _{k=1}^{\infty }{\frac {(-4+1)^{2k}}{k^{2}9^{k}}}\\&=\sum _{k=1}^{\infty }{\frac {(-3)^{2k}}{k^{2}9^{k}}}\\&=\sum _{k=1}^{\infty }{\frac {9^{k}}{k^{2}9^{k}}}\\&=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\\end{aligned}}$ and this converges by the p-series test. Similarly, when $x=2$, we have ${\begin{aligned}\sum _{k=1}^{\infty }a_{k}(2)&=\sum _{k=1}^{\infty }{\frac {(2+1)^{2k}}{k^{2}9^{k}}}\\&=\sum _{k=1}^{\infty }{\frac {(3)^{2k}}{k^{2}9^{k}}}\\&=\sum _{k=1}^{\infty }{\frac {9^{k}}{k^{2}9^{k}}}\\&=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\\end{aligned}}$ and this also converges by the p-series test. Thus, the interval of convergence is $\color {blue}-4\leq x\leq 2$

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