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

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!

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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. 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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