Answer
$\Sigma_{n=1}^\infty nx^{n-1}=\Sigma_{n=0}^\infty (n+1)x^{n}$
Work Step by Step
Since, we know that the Taylor Series for $\dfrac{1}{1-x}$ is defined as:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$
Now, $ (\dfrac{1}{1-x})'=[1+x+x^2+....+x^n]'$
or, $\dfrac{1}{(1-x)^2}=1+2x+3x^2+.....nx^{n-1}$
Thus, we have $=\Sigma_{n=1}^\infty nx^{n-1}=\Sigma_{n=0}^\infty (n+1)x^{n}$
