How to prove that the odd continued fraction approximants of ln(1+X) are upper bounds? The odd order continued fraction approximants for $\ln(1+X)$ are
$$X,\quad \frac{X^2+6X}{4X+6,}\quad \frac{X^3+21X^2+30X}{9X^2+36X+30,}\quad \dots.$$
In "Some bounds for the logarithmic function", Flemming Topsøe remarks that they are also $[n, n − 1]$-Padé approximants. Specific instances such as those shown above can trivially be seen to be upper bounds for $\ln(1+X)$ by taking the difference, differentiating, and observing that the result has the form $X^m/f(X)$ (where $m$ is odd and $f(X)$ is always positive). However, checking a few cases is not the same as proving the result in general. Topsøe presents closed forms for these approximants: rational functions of complicated summations involving multinomial coefficients. He also defines the numerators and denominators recursively. Nevertheless, proving the result by induction looks infeasible due to the sheer complexity of the formulas. Topsøe hints that these results are known to "experts", but his essay is maddeningly short on specifics. Am I overlooking something obvious?
I need to formalise this proof by machine, so some sort of elementary argument would be preferable.
 A: The main result of the paper you link is that
$$\log(1+x) = \frac{x}{1+} \frac{x}{2+} \frac{x}{3+} \frac{2^2 x}{4+} \frac{2^2 x}{5+} \cdots$$
and what you want to know is that truncating this formula at an odd number of steps provides an upper bound. (I am using the standard shorthand $\frac{a}{b+} \frac{c}{d+} \cdots$ to denote $a/(b+c/(d+ \cdots ))$.)
This is a general fact about 
$$\frac{a_1}{b_1+} \frac{a_2}{b_2+} \cdots$$
where $a_i$ and $b_i$ are any positive numbers. Define $c(m,n) = \frac{a_m}{b_m+} \frac{a_{m+1}}{b_{m+1}+} \cdots \frac{a_n}{b_n}$ for any positive $a_i$ and $b_i$. I claim that 
$$c(1,2) < c(1,4) < c(1,6) < \cdots < c(1,5) < c(1,3) < c(1,1). \quad (\ast)$$
We have $\frac{a_{2n-1}}{b_{2n-1}+} \frac{a_{2n}}{b_{2n}} >0$. Now, $\frac{a}{b+y}$ is a decreasing function of $y$, so that gives
$$\begin{array}{rcr}
\frac{a_{2n-2}}{b_{2n-2}+} \frac{a_{2n-1}}{b_{2n-1}+} \frac{a_{2n}}{b_{2n}} &<& \frac{a_{2n-2}}{b_{2n-2}} \\
\frac{a_{2n-3}}{b_{2n-3}+} \frac{a_{2n-2}}{b_{2n-2}+} \frac{a_{2n-1}}{b_{2n-1}+} \frac{a_{2n}}{b_{2n}} &>& \frac{a_{2n-3}}{b_{2n-3}+} \frac{a_{2n-2}}{b_{2n-2}} \\
\frac{a_{2n-4}}{b_{2n-4}+}\frac{a_{2n-3}}{b_{2n-3}+} \frac{a_{2n-2}}{b_{2n-2}+} \frac{a_{2n-1}}{b_{2n-1}+} \frac{a_{2n}}{b_{2n}} &<& \frac{a_{2n-4}}{b_{2n-4}+}\frac{a_{2n-3}}{b_{2n-3}+} \frac{a_{2n-2}}{b_{2n-2}} \\
\end{array}$$
etcetera. So $c(1,2n-2) < c(1,2n)$. A similar argument establishes that $c(1,2n-3) > c(1,2n-1)$ for all $n$, that $c(1,2n) < c(1,2n+1)$ and $c(1,2n+1) > c(1,2n)$. Putting all of that together, we have the inequalities of $(\ast)$. 
So all the odd convergents are upper bounds, and all the even convergents are lower bounds, for whatever the limit is. According to the paper you linked, the limit is $\log(1+x)$, so we win.
