Science:Math Exam Resources/Courses/MATH103/April 2013/Question 01 (f)

MATH103 April 2013

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Question 01 (f)
Sequences and Series: Short Answer Problems - Determine whether the following sequences and series converge (full marks for correct answer with justification; work must be shown for partial marks) Determine all x such that the series $\displaystyle \sum _{n=0}^{\infty }{\frac {n(x+2)^{n}}{3^{n}(n+5)}}$ converges.

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
This type of problem always succumbs to the ratio test.

Hint 2
Don't forget to test the endpoints of your interval where the ratio test is inconclusive!

Hint 3
The endpoints can be shown that when plugged into the series diverge by the divergence test.

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 1

Using the ratio test, we have that ${\begin{aligned}\lim _{n\to \infty }{\big |}{\frac {a_{n+1}}{a_{n}}}{\big |}&=\lim _{n\to \infty }{\big |}a_{n+1}a_{n}^{-1}{\big |}\\&=\lim _{n\to \infty }{\big |}{\frac {(n+1)(x+2)^{n+1}}{3^{n+1}((n+1)+5)}}\cdot {\frac {3^{n}(n+5)}{n(x+2)^{n}}}{\big |}\\&=\lim _{n\to \infty }{\big |}{\frac {(n+1)(x+2)}{3(n+6)}}\cdot {\frac {(n+5)}{n}}{\big |}\\&={\frac {|x+2|}{3}}\lim _{n\to \infty }{\frac {(n+1)(n+5)}{n(n+6)}}\\&={\frac {|x+2|}{3}}\lim _{n\to \infty }{\frac {n^{2}(1+{\tfrac {1}{n}})(1+{\tfrac {5}{n}})}{n^{2}(1+{\tfrac {6}{n}})}}\\&={\frac {|x+2|}{3}}\lim _{n\to \infty }{\frac {(1+{\tfrac {1}{n}})(1+{\tfrac {5}{n}})}{1+{\tfrac {6}{n}}}}\\&={\frac {|x+2|}{3}}\cdot (1)\\&={\frac {|x+2|}{3}}\end{aligned}}$ Now the ratio test tells us that our original series converges when $\displaystyle |x+2|/3<1$. This occurs when $\displaystyle |x+2|<3$ so whenever the distance from 2 is bounded above by 3. This occurs when $\displaystyle -5<x<1$. Another way to see this is to note that $\displaystyle |x+2|<3$ is equivalent to $\displaystyle -3<x+2<3$ and subtracting two from each side gives $\displaystyle -5<x<1$. The ratio test also tells us that the series diverges when $\displaystyle |x+2|>3$ and this occurs when $\displaystyle x+2>3$ or $\displaystyle -(x+2)<3$. Isolating for x yields $\displaystyle x<-5$ or $\displaystyle x>1$. This leaves only the cases when $\displaystyle |x+2|=3$, that is, when $\displaystyle x=-5,1$. To check these, we plug them into the original series. For $\displaystyle x=-5$ we have ${\begin{aligned}\sum _{n=0}^{\infty }{\frac {n(x+2)^{n}}{3^{n}(n+5)}}&=\sum _{n=0}^{\infty }{\frac {n(-5+2)^{n}}{3^{n}(n+5)}}\\&=\sum _{n=0}^{\infty }{\frac {n(-3)^{n}}{3^{n}(n+5)}}\\&=\sum _{n=0}^{\infty }{\frac {n(-1)^{n}3^{n}}{3^{n}(n+5)}}\\&=\sum _{n=0}^{\infty }{\frac {n(-1)^{n}}{n+5}}\end{aligned}}$ Let's look at the limit of the terms. Since as n tends to infinity, we have that the sequence defined by $\displaystyle a_{n}={\frac {n(-1)^{n}}{n+5}}$ has two subsequences that tend to different values (the even indexed terms limit to 1 since the numerator is positive and the odd terms limit to -1 since the numerator is negative). Therefore, the divergence test tells us that the series diverges when $\displaystyle x=-5$ When $\displaystyle x=1$, we have that ${\begin{aligned}\sum _{n=0}^{\infty }{\frac {n(x+2)^{n}}{3^{n}(n+5)}}&=\sum _{n=0}^{\infty }{\frac {n(1+2)^{n}}{3^{n}(n+5)}}\\&=\sum _{n=0}^{\infty }{\frac {n3^{n}}{3^{n}(n+5)}}\\&=\sum _{n=0}^{\infty }{\frac {n}{n+5}}\end{aligned}}$ Let's look at the limit of the terms. Since as n tends to infinity, we have that the sequence defined by $\displaystyle a_{n}={\frac {n}{n+5}}={\frac {1}{1+{\tfrac {5}{n}}}}$ converges to 1 as n tends to infinity. Therefore, the divergence test tells us that the series diverges when $\displaystyle x=1$. Thus the set of x values where this series converges is defined by $\displaystyle -5<x<1$.

Solution 2

This is a second solution for checking the endpoints portion of the question. After finding the radius of convergence as in Solution 1, this leaves only the cases when $\displaystyle |x+2|=3$. To check these (simultaneously), we check this value in the original series with absolute values. We have ${\begin{aligned}\sum _{n=0}^{\infty }{\frac {n|x+2|^{n}}{3^{n}(n+5)}}&=\sum _{n=0}^{\infty }{\frac {n3^{n}}{3^{n}(n+5)}}\\&=\sum _{n=0}^{\infty }{\frac {n}{n+5}}\end{aligned}}$ Let's look at the limit of the terms. Since as n tends to infinity, we have that the sequence defined by $\displaystyle a_{n}={\frac {n}{n+5}}={\frac {1}{1+{\tfrac {5}{n}}}}$ converges to 1 as n tends to infinity. This means the summands $\displaystyle {\frac {n(x+2)^{n}}{3^{n}(n+5)}}$ does not converge to 0 any time $\displaystyle |x+2|=3$. Therefore, the divergence test tells us that the series diverges at the endpoints. Thus the set of x values where this series converges is defined by $\displaystyle -5<x<1$.

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