Science:Math Exam Resources/Courses/MATH101/April 2013/Question 11

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MATH101 April 2013

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Question 11

Full-Solution Problems. In questions 4–12, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. Consider the power series $\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}(x+2)^{n}}{\sqrt {n}}}$, where $\displaystyle x$ is a real number. Find the interval of convergence of this series.

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
Start by finding the radius of convergence.

Hint 2
Use the Ratio Test to calculate the radius of convergence.

Hint 3
Remember to check the endpoints of the interval!

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

We begin by calculating the radius of convergence, using the Ratio Test: $\displaystyle {\begin{aligned}\lim _{n\to \infty }\left|{\frac {(-1)^{n+1}(x+2)^{n+1}}{\sqrt {n+1}}}\cdot {\frac {\sqrt {n}}{(-1)^{n}(x+2)^{n}}}\right|&=\lim _{n\to \infty }{\sqrt {\frac {n}{n+1}}}\cdot |x+2|\\&=\lim _{n\to \infty }{\sqrt {\frac {1}{1+{\tfrac {1}{n}}}}}\cdot |x+2|\\&=|x+2|\end{aligned}}$ By the Ratio Test, the series converges if $\displaystyle |x+2|<1$ and diverges if $\displaystyle |x+2|>1$. Now we need to check the endpoints. When $\displaystyle x+2=-1$, that is, $\displaystyle x=-3$, we have $\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x+2)^{n}}{\sqrt {n}}}&=\sum _{n=1}^{\infty }{\frac {(-1)^{n}(-1)^{n}}{\sqrt {n}}}=\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}\end{aligned}}$ which diverges by the p-series test, since $\displaystyle 1/2<1$. When $\displaystyle x+2=1$, that is, $\displaystyle x=-1$, we have $\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x+2)^{n}}{\sqrt {n}}}&=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}\end{aligned}}$ This is an alternating series with $\displaystyle {\begin{aligned}b_{n}&={\frac {1}{\sqrt {n}}}\end{aligned}}$ Clearly, $\displaystyle b_{n+1}\leq b_{n}$ and $\displaystyle \lim _{n\to \infty }b_{n}=0$. By the Alternating Series Test, $\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x+2)^{n}}{\sqrt {n}}}&=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}\end{aligned}}$ converges. Therefore, the interval of convergence is $\displaystyle (-3,-1]$.

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