A continued fraction involving Bernoulli numbers Let $B_n$ be the Bernoulli numbers. Then, we can write a function $F(x)$, expressed as a continued fraction involving those $B_n$, such that it gives the form, 
$$ \displaystyle    
\displaystyle   F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{B_2+ \cfrac{(3/x)}{B_3+ \cfrac{(4/x)}{B_4+ \cfrac{(5/x)}{B_5+ \cfrac{(6/x)}{B_6+ \cfrac{(7/x)}{B_7+ \cfrac{(8/x)}{\dots}}}}}}}}=\cfrac{x-2}{x}  
$$
Can anyone provide a proof for this finding?
Regards
http://tardigrados.wordpress.com/2013/01/29/numeros-de-bernoulli-y-las-fracciones-continuas-espejo/
 A: Let's start with your definition:
$$F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{B_2+ \cfrac{(3/x)}{B_3+ \cfrac{(4/x)}{B_4+ \cfrac{(5/x)}{B_5+ \cfrac{(6/x)}{B_6+ \cfrac{(7/x)}{B_7+ \cfrac{(8/x)}{\dots}}}}}}}}$$
The only non-zero odd Bernoulli number is $B_1$, so we have
$$F(x)= B_0+ \cfrac{(1/x)}{
  B_1+ \cfrac{(2/x)}{
    B_2+ \frac{3}{4}\left(
      B_4+ \frac{5}{6}\left(
        B_6+ \frac{7}{8}\left(B_8 + \ldots\right)\right)\right)}}$$
Define $K = B_2+ \frac{3}{4}\left(
      B_4+ \frac{5}{6}\left(
        B_6+ \frac{7}{8}\left(B_8 + \ldots\right)\right)\right)$ and we have
$$F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{K}} \
= B_0+ \frac{K}{B_1 Kx+ 2} = 1+ \frac{2K}{4 - Kx} = \frac{4 + (2 - x)K}{4 - Kx} $$
So if $F(x)= \frac{x-2}{x}$ then $$\frac{x-2}{x} = \frac{4 + (2 - x)K}{4 - Kx} \\
(x-2)(4 - Kx) = x(4 + (2 - x)K) \\
4x-8-Kx^2+2Kx = 4x + 2Kx-Kx^2 \\
-8 = 0$$
This might be why you have had problems finding a proof, but it might also hint at how your identity can be corrected.
Addendum: actually, $K$ seems to diverge quite fast, so you might be able to take limits and show that as the number of terms increases the partial continued fraction $\frac{4 + (2 - x)k_n}{4 - k_nx}$ tends to $\frac{2-x}{-x}$.
