Finding an Expression as an Elementary Function for a Power Series

Question

Consider $$h(z)=\sum_{n=1}^{\infty}\frac{(z-2)^n}{n}.$$ I wish to find an expression for $h$ as an elementary function.

This question has me stumped. I considered another function, $$f(z)=\sum_{n=1}^{\infty} n(z-2)^n.$$ This is much easier to express as an elementary function, as $$f(z)=\sum_{n=1}^{\infty} n(z-2)^n=(z-2)\frac{d}{dz}\sum_{n=1}^{\infty} (z-2)^n=\frac{z+3}{(z+2)^2}.$$ But for the function $h$, I cannot see a similar technique or a manipulation to yield such a function.

I would really appreciate a hint.

Answer 1

Hint: formally $h'(z)=\sum (z-2)^{n-1}$. Calculate this sum and integrate. The answer is $Log (3-z)$ for $|z-2| <1$ where $Log$ is the principle branch of logarithm. You have to have some knowledge of logarithms in the complex plane to answer this question.

Answer 2

Hint:

Instead of looking at $h$, first try to calculate what $$\frac{\partial h}{\partial z}$$ would be.

Answer 3

Taking the convergent part of this sum and making

$$ y = z-2 $$

we have

$$ h(y) = \sum_{k=1}^{\infty}\frac{y^k}{k} = \int_0^y\sum_{k=0}^{\infty}\zeta^kd\zeta = \int_0^y\frac{d\zeta}{1-\zeta} $$