<s>[INST] Logarithm of the hypergeometric function For $F(x)={}_2F_1 (a,b;c;x)$, with $c=a+b$, $a>0$, $b>0$, it has been proved in [1] that $\log F(x)$ is convex on $(0,1)$.
I numerically checked that with a variety of $a,\ b$ values, $\log F(x)$ is not only convex, but also has a Taylor series in x consisting of strictly positive coefficients. Can this be proved?
[1] Generalized convexity and inequalities, Anderson, Vamanamurthy, Vuorinen, Journal of Mathematical Analysis and Applications, Volume 335, Issue 2,
http://www.sciencedirect.com/science/article/pii/S0022247X07001825#
 [/INST] Here's a sketch of a proof of a stronger statement: the coefficients of the Taylor series for $\log{}_2F_1(a,b;a+b+c;x)$ are rational functions of $a$, $b$, and $c$ with positive coefficients.
To see this we first note that 
$$\begin{aligned}
\frac{d\ }{dx} \log {}_2F_1(a,b;a+b+c;x) &= 
\frac{\displaystyle
\frac{d\ }{dx}\,{}_2F_1(a,b;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}\\[3pt]
 &=\frac{ab}{a+b+c}\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b;a+b+c;x)}.
\end{aligned}
$$
Then 
$$
\begin{gathered}
\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b;a+b+c;x)}
  = \frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b+1;a+b+c;x)} \\
 \hfill\times
  \frac{{}_2F_1(a,b+1;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}.\quad
\end{gathered}
$$
We have continued fractions for the two quotients on the right.
Let $S(x; a_1, a_2, a_3, \dots)$ denote the continued fraction
$$\cfrac{1}{1-\cfrac{a_1x}
{1-\cfrac{a_2x}
{1-\cfrac{a_3x}
{1-\ddots}
}}}
$$
Then
$$\begin{gathered}\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b+1;a+b+c;x)}
 = S \left( x;{\frac { \left( b+1 \right)  \left( b+c \right) }{ \left( a
+b+c+1 \right)  \left( a+b+c \right) }},
\right.\hfill\\
\left.
{\frac { \left( a+1 \right) 
 \left( a+c \right) }{ \left( a+b+c+2 \right)  \left( a+b+c+1 \right) 
}}, 
{\frac { \left( b+2 \right)  \left( b+c+1 \right) }{ \left( a+b+c+3
 \right)  \left( a+b+c+2 \right) }},
 \right.\\
\hfill
\left.
 {\frac { \left( a+2 \right) 
 \left( a+c+1 \right) }{ \left( a+b+c+4 \right)  \left( a+b+c+3
 \right) }},\dots \right) 
 \end{gathered}
$$
and 
$$\begin{gathered}
\frac{{}_2F_1(a,b+1;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}
=S \left( x,{\frac {a}{a+b+c}},
{\frac { \left( b+1 \right)  \left( b+c
 \right) }{ \left( a+b+c+1 \right)  \left( a+b+c \right) }},
 \right.\hfill\\
 \left.
 {\frac {
 \left( a+1 \right)  \left( a+c \right) }{ \left( a+b+c+2 \right) 
 \left( a+b+c+1 \right) }},
 {\frac { \left( b+2 \right)  \left( b+c+1
 \right) }{ \left( a+b+c+3 \right)  \left( a+b+c+2 \right) }},
  \right.\\
 \hfill
 \left.
 {\frac {
 \left( a+2 \right)  \left( a+c+1 \right) }{ \left( a+b+c+4 \right) 
 \left( a+b+c+3 \right) }}, \dots\right) 
\end{gathered}
$$
The first of these continued fractions is Gauss's well-known continued fraction, and the second can easily be derived from the first. It follows from these formulas that the coefficients of the Taylor series for $\log{}_2F_1(a,b;a+b+c;x)$ are rational functions of $a$, $b$, and $c$ with positive coefficients.
</s>