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

From the question we have $$\sum _{n=0}^{\mathrm{\infty }}n{x}^{n+1}=x^2\sum _{n=0}^{\mathrm{\infty }}n{x}^{n-1}.$$

Therefore, it is enough to find an explicit expression of $\sum _{n=0}^{\mathrm{\infty }}n{x}^{n-1}$. For simplicity we denote $f(x)={\sum }_{n=0}^{\mathrm{\infty }}n{x}^{n-1}$.

Observe that $n{x}^{n-1}=\frac{d}{dx}{x}^n$. Using this, $f$ can be written as $$f(x)=\sum _{n=0}^{\mathrm{\infty }}n{x}^{n-1}=\sum _{n=0}^{\mathrm{\infty }}\frac{d}{dx}{x}^n.$$

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 $$f(x)=\frac{d}{dx}\left(\sum _{n=0}^{\mathrm{\infty }}{x}^n\right).$$

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 ${\sum }_{n=0}^{\mathrm{\infty }}n{x}^{n+1}=x^{2}f(x)=\frac{x^2}{(1-x{)}^2}.$

Answer: ${\sum }_{n=0}^{\mathrm{\infty }}n{x}^{n+1}=\frac{x^2}{(1-x{)}^2}.$