Science:Math Exam Resources/Courses/MATH101/April 2014/Question 09 (b)

MATH101 April 2014

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[hide] Question 09 (b)
Long Problem. Show your work. No credit will be given for the answer without the correct accompanying work. Express $\sum _{n=0}^{\infty }n^{2}x^{n}$ as a ratio of polynomials. For which x does this series converge?

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.

[show] Hint 1
Use the formula in (a). Check the derivative of it.

[show] Hint 2
Use the ratio test for the interval of convergence. Check the boundaries 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.

[show] Solution

Start with the formula we derived in (a). If we differentiate this formula with respect to $x$, ${\begin{aligned}\sum _{n=0}^{\infty }n^{2}x^{n-1}&={\frac {d}{dx}}\sum _{n=0}^{\infty }nx^{n}\\&={\frac {d}{dx}}\left({\frac {x}{(1-x)^{2}}}\right)\\&={\frac {1}{(1-x)^{2}}}+{\frac {2x}{(1-x)^{3}}}\\&={\frac {1+x}{(1-x)^{3}}}.\end{aligned}}$ If we multiply both sides by $x$: ${\begin{aligned}\sum _{n=0}^{\infty }n^{2}x^{n}={\frac {x(1+x)}{(1-x)^{3}}}\end{aligned}}$ then we have the series we are interested in. We note that the two expression are only equivalent if the series converges. We will check the radius of convergence by using the ratio test. ${\begin{aligned}\lim _{n\to \infty }\left|{\frac {(n+1)^{2}x^{n}}{n^{2}x^{n-1}}}\right|=\left(\lim _{n\to \infty }\left|{\frac {(n+1)^{2}}{n^{2}}}\right|\right)|x|=|x|.\end{aligned}}$ By the ratio test, if $|x|<1$, the series converges and if $|x|>1$, the series diverges. This is unsurprising since we already know this is required for the geometric series in (a). We have to check the boundaries independently. If $x=1$, the expression ${\frac {x(1+x)}{(1-x)^{3}}}$ fails to exist and so the series must diverge. If $x=-1$, this expression is finite but that does not mean the series converges. We have to look at the series itself which becomes ${\begin{aligned}\sum _{n=0}^{\infty }(-1)^{n}n^{2}&=-1^{2}+2^{2}-3^{2}+4^{2}-5^{2}+6^{2}+\cdots =(-1^{2}+2^{2})+(-3^{2}+4^{2})+(-5^{2}+6^{2})+\cdots \\&=(2-1)(1+2)+(3-2)(3+4)+(6-5)(5+6)+\cdots =\sum _{n=0}^{\infty }n,\end{aligned}}$ and hence diverges. Thus, the series only converges for $x\in (-1,1)$.