Science:Math Exam Resources/Courses/MATH101/April 2014/Question 01 (g)

MATH101 April 2014

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q2  •  Q3  •  Q4  •  Q5  •  Q6  •  Q7 (a)  •  Q7 (b)  •  Q8  •  Q9 (a)  •  Q9 (b)  •  

   Other MATH101 Exams

  •  April 2011  •  April 2010  •  April 2009  •  April 2008  •  April 2007  •  April 2005  •  April 2006  •  April 2012  •  April 2013  •  April 2014  •  April 2015  •  April 2016  •  April 2018  •  April 2017  •  

Question 01 (g)
Short Problem. Show your work. No credit will be given for the answer without the correct accompanying work. Find the radius of convergence for the power series $\sum _{n=0}^{\infty }{\frac {(x-2)^{n}}{n^{2}+1}}$.

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 using the ratio test.

Hint 2
The ratio test applies to series of the form $\sum _{n=0}^{\infty }a_{n}$ and states that if $R=\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|$ then: If $R<1$, $\sum _{n=0}^{\infty }a_{n}$ is absolutely convergent If $R>1$, $\sum _{n=0}^{\infty }a_{n}$ is divergent If $R=1$, the ratio test is inconclusive and so we must use another test to determine convergence

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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Applying the ratio test with $a_{n}={\frac {(x-2)^{n}}{n^{2}+1}}$, we find that ${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {(x-2)^{n+1}}{(n+1)^{2}+1}}\cdot {\frac {n^{2}+1}{(x-2)^{n}}}\right|\\&=\lim _{n\to \infty }|x-2|\left|{\frac {n^{2}+1}{n^{2}+2n+2}}\right|\\&=|x-2|\lim _{n\to \infty }\left|{\frac {1+{\frac {1}{n^{2}}}}{1+{\frac {2}{n}}+{\frac {2}{n^{2}}}}}\right|\\&=|x-2|.\end{aligned}}$ The radius of convergence is the bound on $x$ values that allows the series to converge. Using the ratio test, the series converges if the ratio tends to a number smaller than 1. Hence the series $\sum _{n=0}^{\infty }{\frac {(x-2)^{n}}{n^{2}+1}}$ converges for $\displaystyle |x-2|<{\color {blue}1}$ by the ratio test, and so its radius of convergence is 1.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Power series, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag