Science:Math Exam Resources/Courses/MATH101/April 2017/Question 11 (c)

MATH101 April 2017

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Question 11 (c)
Let ${A_{n}}$ be a sequence of positive numbers such that the power series $\sum _{n=0}^{\infty }A_{n}(x-1)^{n}$ has radius of convergence $R={\frac {3}{2}}$. (c) Does the series $\sum _{n=0}^{\infty }n(n-1)A_{n}$ converge or diverge? Justify your answer.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that when a power series $\sum _{n=0}^{\infty }a_{n}(x-c)^{n}$ has a radius of convergence $R$, its derivative has the same radius of convergence.

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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. Let $f(x)=\sum _{n=0}^{\infty }A_{n}(x-1)^{n}$. Then, we can observe that $f''(x)=(\sum _{n=0}^{\infty }A_{n}(x-1)^{n})''=(\sum _{n=1}^{\infty }nA_{n}(x-1)^{n-1})'=\sum _{n=2}^{\infty }n(n-1)A_{n}(x-1)^{n-2}$. This implies that $f''(2)=\sum _{n=2}^{\infty }n(n-1)A_{n}=\sum _{n=0}^{\infty }n(n-1)A_{n}$. The last equality follows from $n(n-1)=0$ for $n=0$ and $n=1$. By the Hint, taking derivative doesn't change the radius of convergence, so that $f''(x)$ also has the radius of convergence ${\frac {3}{2}}$. Since $|2-1|=1<{\frac {3}{2}}$, the series $f''(2)=\sum _{n=0}^{\infty }n(n-1)A_{n}$ converges. Answer: $\color {blue}\sum _{n=0}^{\infty }n(n-1)A_{n}<+\infty$