How do I prove the partial denominators formula of the Bauer-Muir transformation of a generalized continued fraction? Notation: $b_{0}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( a_{n}/b_{n}\right) $ is the Gauss Notation for generalized continued fractions.

Description of the Bauer-Muir transformation 
(Based on pp. 76-77 of Lisa Lorentzen and Haakon Waadeland's book (A), chapter II, theorems 11, and Theorem 7;  also section 5 of J. Mc Laughlin, and Nancy J. Wyshinski's online paper (C), and section 5 of Bruce C. Berndt, Sen-Shan Huang, Jaebum Sohn, and Seung Hwan Son's online paper (B) ). 
Given a convergent c.f. $\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})=\lim_{n\rightarrow \infty }A_{n}/B_{n}$ and a sequence ${w_{n}}$ we can construct a new c.f $\underset{n=1}{\overset{\infty }{\mathbb{K}}}({c_{n}/d_{n}})=\lim_{n\rightarrow \infty }C_{n}/D_{n}$ (if convergent) which is its Bauer-Muir transform with respect to ${w_{n}}$. 
By theorem 11, chapter II, of Lisa Lorentzen and Haakon Waadeland's book (A), pp.76-77, the relations between $A_{n},B_{n}$ and $C_{n},D_{n}$ are given by:
$C_{n}=A_{n}+A_{n-1}w_{n}$, $D_{n}=B_{n}+B_{n-1}w_{n}$, with the initial conditions $C_{-1}=1,D_{-1}=0.$
If for $n\geq 1$,  $\lambda_{n}=a_{n}-w_{n-1}(b_{n}+w_{n})\neq 0$, then
$$c_{n}=a_{n-1}\lambda_{n}/\lambda_{n-1},$$ 
$$d_{n}=b_{n}+w_{n}-w_{n-2}\lambda_{n}/\lambda _{n-1},$$
for $n\geq 2,$ and
$$\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})=w_{0}+\dfrac{\lambda_{1}}{b_{1}+w_{1}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}(c_{n}/d_{n})}.$$
The elements of $\underset{n=1}{\overset{\infty }{\mathbb{K}}}(a_{n}/b_{n})$ are computed by an application of Theorem 7, chapter II, of Lisa Lorentzen and Haakon Waadeland's book (A), which transforms a sequence into a continued fraction. I was able to derive $c_{n}$ but not $d_{n}$. 

Example: Application to the $\log (1+t)$ expansion. By choosing $w_{0}=w_{1}=0,w_{n}=(n-1)t$, for $n\geq 2$, one can derive
$$\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( n^{2}t/\left( \left(
n+1\right) -kt\right) \right) =\dfrac{t}{2+\underset{n=3}{\overset{\infty }{%
\mathbb{K}}}\left( \left( n-2\right) ^{2}t/\left( n-\left( n-3\right)
t\right) \right) }.$$
Hence (see Wikipedia) the expansion
$$\log (1+t)=\displaystyle\sum_{n=1}^{\infty }\dfrac{(-1)^{n-1}t^{n}}{n}=\dfrac{t}{1+%
\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( n^{2}t/\left( \left(
n+1\right) -nt\right) \right) },$$
can be improved with respect to the convergence speed to this one 
$$\log (1+t)=\dfrac{t}{1+\dfrac{t}{2+\underset{n=3}{\overset{\infty }{\mathbb{%
K}}}\left( \left( n-2\right) ^{2}t/\left( n-\left( n-3\right) t\right)
\right) }}.$$

Derivation of $c_n$
Here is how I got $c_n$. In page 77 of reference A it is proved that
$$C_{n-1}D_n-D_{n-1}C_n=(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\lambda_n.$$ 
Hence
$$C_{n-2}D_{n-1}-D_{n-2}C_{n-1}=(A_{n-2}B_{n-3}-A_{n-3}B_{n-2})\lambda_{n-1}.$$ 
For $n\ge 2$ Theorem 7 of reference A (derived from the fundamental c.f. recurrence) gives  
$$c_n=\dfrac{C_{n-1}D_n-D_{n-1}C_n}{-(C_{n-2}D_{n-1}-D_{n-2}C_{n-1})}=\dfrac{(A_{n-1}B_{n-2}-A_{n-2}B_{n-1})\lambda_n}{-(A_{n-2}B_{n-3}-A_{n-3}B_{n-2})\lambda_{n-1}}.$$
By the determinant formula we have
$$A_{n-1}B_{n-2}-A_{n-2}B_{n-1}=-a_{n-1}(A_{n-2}B_{n-3}-A_{n-3}B_{n-2}).$$
Thus
$$c_n=a_{n-1}\dfrac{\lambda_n}{\lambda_{n-1}}.$$


QUESTION 1: How does one prove 
$$d_{n}=b_{n}+w_{n}-w_{n-2}\lambda_{n}/\lambda_{n-1}$$
from 
$$d_{n}=\dfrac{C_{n}D_{n-2}-D_{n}C_{n-2}}{C_{n}D_{n-1}-D_{n}C_{n-1}}\qquad\text{for}\quad n\ge 2$$
and
$\lambda_{n}=a_{n}-w_{n-1}(b_{n}+w_{n})$,
  $C_{n}=A_{n}+A_{n-1}w_{n}$,
  $D_{n}=B_{n}+B_{n-1}w_{n}$ ?


References
A - Lisa Lorentzen and Haakon Waadeland, Continued Fractions and Applications, North-Holland, Amsterdam, 1992. (pdf file of pp. 76-77) 
B - Bruce C. Berndt, Sen-Shan Huang, Jaebum Sohn, and Seung Hwan Son, A Transformation Formula in Rogers--Ramanujan Continued Fraction. (section 5) 
C - J. Mc Laughlin, and Nancy J. Wyshinski, Real numbers with polynomial continued fraction expansions, arXiv, 2004. 
 A: Answer to Question 1: 
Using the following relations (in Perron, Die Lehre von den Kettenbrüchen, Band II, 1957 and Sergey Khrushchev, Orthogonal Polynomials and
Continued Fractions: From Euler's Point of View, 2008) 
$$\begin{equation}
A_{n}B_{n-1}-A_{n-1}B_{n}=\left( -1\right) ^{n-1}a_{1}a_{2}\cdots a_{n} 
\nonumber
\end{equation}$$
$$\begin{equation}
A_{n}B_{n-2}-A_{n-2}B_{n}=\left( -1\right) ^{n}b_{n}a_{1}a_{2}\cdots a_{n-1}
\end{equation}$$
$$\begin{equation}
A_{n}B_{n-3}-A_{n-3}B_{n }=(-1)^{n-1}a_{1}a_{2}\cdots a_{n -2}\left(
b_{n}b_{n-1}+a_{n}\right)   
\end{equation}$$
we can derive
$$\begin{eqnarray}
&&\left( A_{n}+w_{n}A_{n-1}\right) \left( B_{n-1}+w_{n-1}B_{n-2}\right)   \\
&&-\left( A_{n-1}+w_{n-1}A_{n-2}\right) \left( B_{n}+w_{n}B_{n-1}\right)  \\
&=&-\left( -1\right) ^{n }a_{1}a_{2}\cdots a_{n-1}\left(
a_{v}-w_{n-1}\left( b_{n}+w_{n}\right) \right)   
\end{eqnarray}$$
and
$$\begin{eqnarray}
&&\left( A_{n -1}+w_{n-1}A_{n-2}\right) \left(
B_{n-2}+w_{n-2}B_{n-3}\right)   \nonumber \\
&&-\left( A_{n-3}+w_{n-3}A_{\nu -3}\right) \left( B_{n-1}+w_{n-1}B_{n
-2}\right)    \\
&=&-\left( -1\right) ^{n -1}a_{1}a_{2}\cdots a_{n-2}\left(
a_{n-1}-w_{n-2}\left( b_{n-1}+w_{n-1}\right) \right)   
\end{eqnarray}$$
as well as
$$\begin{eqnarray}
&&\left( A_{n}+w_{n}A_{n-1}\right) (B_{n-2}+w_{n-2}B_{n-3})   \\
&&-(A_{n-2}+w_{n-2}A_{n-3})\left( B_{n}+w_{n}B_{n -1}\right)    \\
&=&\left( -1\right) ^{n}b_{n }a_{1}a_{2}\cdots a_{n -1}+w_{n
-2}(-1)^{n -1}a_{1}a_{2}\cdots a_{n -2}\left( b_{n}b_{n-1}+a_{n}\right) 
 \\
&&+w_{n}\left( -1\right) ^{n -2}a_{1}a_{2}\cdots a_{n-1}+w_{n-2}\left(
-1\right) ^{n-1}b_{n-1}a_{1}a_{2}\cdots a_{n-2} 
\end{eqnarray}$$
Therefore
$$\begin{eqnarray*}
d_{n} &=&\frac{\left( A_{n}+w_{n}A_{n-1}\right)
(B_{n-2}+w_{n-2}B_{n-3})-(A_{n-2}+w_{n-2}A_{n-3})\left(
B_{n}+w_{n}B_{n-1}\right) }{%
(A_{n-1}+w_{n-1}A_{n-2})(B_{n-2}+w_{n-2}B_{n-3})-(A_{n-2}+w_{n-2}A_{n-3})(B_{n-1}+w_{n-1}B_{n-2})%
} \\
&=&\frac{\left( -1\right) ^{n}b_{n}a_{1}a_{2}\cdots
a_{n-1}+w_{n-2}(-1)^{n-1}a_{1}a_{2}\cdots a_{n-2}\left(
b_{n}b_{n-1}+a_{n}\right) }{-\left( -1\right) ^{n -1}a_{1}a_{2}\cdots
a_{n-2}\left( a_{n-1}-w_{n-2}\left( b_{n -1}+w_{n-1}\right) \right) } \\
&&+\frac{w_{n }\left( -1\right) ^{n-2}a_{1}a_{2}\cdots
a_{n-1}+w_{n-2}\left( -1\right) ^{n-1}b_{n -1}a_{1}a_{2}\cdots a_{n-2}}{%
-\left( -1\right) ^{n-1}a_{1}a_{2}\cdots a_{n-2}\left( a_{v-1}-w_{n-2}\left(
b_{n-1}+r_{n-1}\right) \right) } \\
&=&\dfrac{\left( b_{n}+w_{n}\right) \left(
a_{n-1}-w_{n-2}(b_{n-1}+w_{n-1})\right) -w_{n-2}\left(
a_{n}-w_{n-1}(b_{n}+w_{n})\right) }{a_{n-1}-w_{n-2}(b_{n-1}+w_{n-1})} \\
&=&b_{n }+w_{n}-w_{n-2}\dfrac{a_{n }-w_{n-1}(b_{n }+w_{n})}{a_{n
-1}-w_{n-2}(b_{n-1}+w_{n-1})}
\end{eqnarray*}$$
This transformation appears also in "Die Transformation von Bauer und Muir", §7 of Oskar Perron, Die Lehre von den Kettenbrüchen, Band II, 1957.
