The question is as follows:
The power series $$\sum _{ k=0 }^{ \infty }{ k{ z }^{ k } } $$ represents an analytic function $f$ in its disk of convergence, the disk of radius $R=1$, centered at the origin. Find $f$.
The answer is:
$$f\left( z \right) =\frac { z }{ { (1-z) }^{ 2 } } $$
I proceeded by a hand-crafted method, ie. I wrote down the beginning of $\sum _{ k=0 }^{ \infty }{ k{ z }^{ k } } $, and just below the $\sum _{ k=0 }^{ \infty }{ { z }^{ k } } $ and take the sum of them term by term. It yielded into an expression already known:
$$f(z) (= \sum _{ k=0 }^{ \infty }{ k{ z }^{ k } }) = \sum _{ k=0 }^{ \infty }{ (k+1){ z }^{ k } } - \sum _{ k=0 }^{ \infty }{ { z }^{ k } } $$
As,
$$ \sum _{ k=0 }^{ \infty }{ { z }^{ k } } =\frac { 1 }{ 1-z } \quad (1)$$ and $$ \sum _{ k=0 }^{ \infty }{ (k+1){ z }^{ k } } =\frac { 1 }{ { (1-z) }^{ 2 } } \quad (2)$$
I reached the solution.
But I feel like cheating, because the conclusion about $(2)$ was evaluated in the course; also my instructor used another method for this question: She differentiated the power series and manipulated analyticity. I also tried to integrate first, then differentiate by regrouping terms, but neither could I achieve nor could I understand the professor's method.
I'd much liked you help me to find alternative solution -as my professor did- to this?
