Science:Math Exam Resources/Courses/MATH101/April 2017/Question 03 (b)/Solution 1

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: ${\textstyle \color {blue}\sum _{n=0}^{\infty }nx^{n+1}={\frac {x^{2}}{(1-x)^{2}}}.}$