Deducing a function from a power series $\in\mathbb{C}$

Question

We're told that some function, $f(z)$, has the power series representation

$$f(z)=\sum_{n=0}^\infty nz^n$$

and told that it has a radius of convergence of $1$, and is analytic within this disc. From this, we are told to find and equation for $f(z)$, as well as evaluating $f\left(\frac{1}{3}\right)$.

I wasn't entirely sure how to go about this, so here is what I've done so far: since we know that all power series' are of the form

$$\sum_{n=0}^\infty a_n(z-z_0)^n$$

where

$$a_n=\frac{f^{(n)}(z_o)}{n!}$$

and what this implies is that

$$n=\frac{f^{(n)}(0)}{n!}$$ $$f^{(n)}(0)=n\cdot n!$$

now, from here, I though that setting up some equalities using the power series representation of $f(z)$ and the derivatives of $f(z)$ at $z=0$ might prove useful, although this method didn't get me very far. I know what the answer is, for reference, it's

$$f(z)=\frac{z}{(z-1)^2}$$

but again, I'm not sure I would arrive at this result. In terms of level, I know of only the surface of complex analysis.

Any responses are appreciated, thank you.

Answer 1

Note that

$$f(z)=\sum_{n=0}^\infty nz^n = \sum_{n=1}^\infty nz^n = z\sum_{n=1}^\infty nz^{n-1}$$

Now, $nz^{n-1} = (z^n)'$, and therefore

$$f(z) = z\sum_{n=1}^\infty nz^{n-1} = z\sum_{n=1}^\infty (z^n)' = z\left(\sum_{n=1}^\infty z^n\right)'$$

It's enough then to find $\sum_{n=1}^\infty z^n$. Can you do it?

Answer 2

Consider the geometric series

$$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$

Taking the derivative gives

$$\frac{1}{(1-z)^2} = \sum_{n=1}^\infty nz^{n-1}$$

Multiplying by $z$ finally gives

$$\frac{z}{(1-z)^2} = \sum_{n=1}^\infty nz^n = f(z)$$

Answer 3

An alternative approach. Observe that $$ f(z)=\sum_{n=1}^\infty nz^n=\sum_{n=1}^\infty\sum_{k=1}^nz^n=\sum_{k=1}^\infty\sum_{n=k}^\infty z^n=\sum_{k=1}^\infty\frac{z^k}{1-z}=\frac{1}{1-z}\frac{z}{1-z}=\frac{z}{(1-z)^2} $$ where the interchanging of summation is allowed since the series converges absolutely on the disc $|z|<1$.