Proving that $\frac{\sum_{n=0}^\infty a_n n z^n}{\sum_{n=0}^\infty a_n z^n}$ is strictly increasing. Im considering the quotient from the title:
$$g(z)=\frac{\sum_{n=0}^\infty n\cdot a_n z^n}{\sum_{n=0}^\infty a_n z^n}$$
with $a_n>0$ $\forall n\in \mathbb{N}$ and for all $z\in(0,1)$. And assuming that both sums suffice the requirements necessary in order to differentiate, etc. What I want to prove is that $g'(z)>0$ and $g''(z)\geq 0$ (Im pretty sure about the truth of the second identity but not 100% sure). 
I've been able to prove the first identity deriving with respect to z and then using the Cauchy formula for the coefficient of the product of two infinite series. (If requested I can post it but it's way too cumbersome to be useful to prove that $g''(z)\geq 0$). I also think that there must be an easier and smarter way to prove it (In fact it may be trivial but I just haven't thought of it).
I would appreciate any help and thanks in advance.
Edit: I've asked this because I think it could be a general property, but the problem this actually comes from is with this specific $a_n$ (if it can be proven for this case I'd be happy too)
$$a_n = \frac{\Gamma\left(\frac{1}{2} +\frac{1}{k}\right)}{\Gamma\left(\frac{1}{2} \right) \cdot \Gamma\left(1 +\frac{1}{k}\right)}\cdot \frac{\Gamma\left(\frac{1}{2} + n \right) \Gamma\left(1+\frac{1}{k} + n \right)}{\Gamma\left(\frac{1}{2} +\frac{1}{k}+ n \right)} \cdot \frac{1}{n!}$$
where $k>0$ is a positive constant. And this problems in itself comes from trying to prove that both the first and second derivative of the following quotient are positive:
$$\frac{_2F_1\left(\frac{3}{2},1+\frac{1}{k};\frac{1}{2}+\frac{1}{k};z \right) }{_2F_1\left(\frac{3}{2},1+\frac{1}{k};\frac{1}{2}+\frac{1}{k};z \right) }$$

Even when the inequality changed to $g''(z) \ge 0$, it need not be true.
For an counterexample, take $a_n = \begin{cases}\frac1{n!},& n \ne 1,\\ \frac12 & n = 1\end{cases}$. The denominator of this $g(z)$ is
$$D(z) \stackrel{def}{=} \sum_{n=0}^\infty a_n z^n = e^z - \frac{z}{2}$$
From this, we can deduce
$$g(z) = \frac{\sum\limits_{n=0}^\infty n a_n z^n}{\sum\limits_{n=0}^\infty a_n z^n} = z\frac{d}{dz}\log(D(z)) = z \frac{e^z - \frac12}{e^z - \frac{z}{2}}$$
If one throw $g(z)$ to a CAS and make a plot of $g''(z)$, one will notice $g''(z) < 0$ for $z \in [1, \sim\!3.81787380305645 ]$.
