Using termwise differentiation and integration find the sum $\sum\limits_{n=0}^{\infty}(n+1)x^n = 1 + 2x + 3x^2 + ... +(n+1)x^n$ Using termwise differentiation and integration find the sum $$\sum\limits_{n=0}^{\infty}(n+1)x^n = 1 + 2x + 3x^2 + \cdots +(n+1)x^n$$
The answer can be achevied simply by setting $S = \sum\limits_{n=0}^\infty (n+1)x^n$ and evaluating $S(1-x) = \sum\limits_{n=1}^\infty x^n = \frac{1}{1-x}$. Hence $S = \frac{1}{(1-x)^2}$. However, this approach does not use termwise approach:
$$f(x) = 1 + 2x + 3x^2 + \cdots + (n+1)x^n$$
$$f'(x) = 2 + 6x + \cdots + n(n+1)x^{n-1}$$
Not sure how to proceed after this point using that approach. Could someone provide a solution? Thanks.

We have
$$\sum_{n\geq 0}(n+1)x^n=\left(\sum_{n\geq 0}x^{n+1}\right)'=\left(\frac{x}{1-x}\right)'=\frac{1}{(1-x)^2}.$$
