Science:Math Exam Resources/Courses/MATH101/April 2017/Question 03 (b)

MATH101 April 2017

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Question 03 (b)
The interval of convergence of the power series $\sum _{n=0}^{\infty }nx^{n+1}$ is $(-1,1)$. (You don’t have to show this.) Find a formula (not involving infinite series) for $\sum _{n=0}^{\infty }nx^{n+1}$, valid for all $-1<x<1$.

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 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 $\sum _{n=0}^{\infty }x^{n}={\frac {1}{1-x}}$ when $x\in (-1,1)$.

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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. From the question we have $$\sum _{n=0}^{\infty }nx^{n+1}=x^{2}\sum _{n=0}^{\infty }nx^{n-1}.$$ Therefore, it is enough to find an explicit expression of $\sum _{n=0}^{\infty }nx^{n-1}$. For simplicity we denote ${\textstyle f(x)=\sum _{n=0}^{\infty }nx^{n-1}}$. Observe that $nx^{n-1}={\frac {d}{dx}}x^{n}$. Using this, $f$ can be written as $$f(x)=\sum _{n=0}^{\infty }nx^{n-1}=\sum _{n=0}^{\infty }{\frac {d}{dx}}x^{n}.$$ Since the interval of convergence of the power series is $(-1,1),$ we can reverse the order of summation and derivative on this interval to get $$f(x)={\frac {d}{dx}}\left(\sum _{n=0}^{\infty }x^{n}\right).$$ By the hint (which can be easily obtained from the explicit expression of a geometric series), this implies that $$f(x)={\frac {d}{dx}}\left({\frac {1}{1-x}}\right).$$ Therefore, computing the derivative based on the chain rule and power rule, the explicit formula for $f$ is given by $$f(x)={\frac {d}{dx}}\left({\frac {1}{1-x}}\right)={\frac {1}{(1-x)^{2}}}.$$ Combining with the first equation, we get ${\textstyle \sum _{n=0}^{\infty }nx^{n+1}=x^{2}f(x)={\frac {x^{2}}{(1-x)^{2}}}.}$ Answer: ${\textstyle \color {blue}\sum _{n=0}^{\infty }nx^{n+1}={\frac {x^{2}}{(1-x)^{2}}}.}$