# 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

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Can someone explain why this question is being downvoted? –  Will Orrick May 3 '13 at 13:19
A -perhaps- nicer formulation is this in terms of zetas at nonpositive integer arguments: $\text{cf}(0;[\zeta(0),\zeta(-1),\zeta(-2),...],[x,x,x,x,...]) = -2 x$ –  Gottfried Helms Jun 4 at 21:17

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 $\frac{2-x}{-x}$.

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Your addendum seems to be ok, but now the next step is to prove that K diverges, if you want a complete proof. –  Albert Zotkin May 4 '13 at 8:51