# Evaluation of a class of continued fractions

Is there a closed-form way of writing the continued fraction: $$1 + \frac{2}{3+ \frac{4}{5 + \frac{6}{7 + ...}}}$$ EDIT: Since the above has been determined as $\frac{1}{\sqrt{e}-1}$, is there a similar expression for: $$2 + \frac{3}{4+ \frac{5}{6 + \frac{7}{8 + ...}}}$$ More generally, are there general closed-form expressions for all continued fractions of the form: $$a_n = n + \frac{n+1}{(n+2) + \frac{n+3}{(n+4) + ...}} \\ f(x) = x + \frac{x+1}{(x+2) + \frac{x+3}{(x+4) + \cdots}} = x + \frac{x+1}{f(x+2)} \\ f(x) f(x+2) = xf(x+2) + x+1$$ And can said closed form be extended to all real numbers? For example, I experimented with extending the sequence to negative values of n and found that for all negative odd $n, a_n = -1$.

• You can find your continued fraction here. – Lucian Aug 31 '14 at 22:42
• A derivation can be found in this paper by Leonhard Euler. See page 14. – Winther Aug 31 '14 at 22:46
• A more general formula can be found here. – Lucian Aug 31 '14 at 23:00
• Perhaps you find this go.helms-net.de/math/divers/GenContFracRationalE.htm interesting, although the sequences of coefficients in your CF goe in steps of 2 and I've a table of CF's with sequences going in steps of 1. But perhaps the table as such is of interest and gives an idea for the analoguous table for your version – Gottfried Helms Sep 3 '14 at 20:48

It's $1/(e^{1/2}-1)$. You should be able to derive this by doing a term-by-term transformation on an appropriate infinite series.
• Could you clarify how you did this? Also, how would you evaluate $a_0$ and more generally $a_n$? – James Harrison Sep 1 '14 at 3:31
If $$z_1 = 1 + 2/(3 + 4/(5 + 6/( ... = {1\over \exp(1/2)-1 }$$ (as given in the answer before) and $$z_3 =3 + 4/(5 + 6/( ...$$ then $$z_1 = 1 + 2/z_3$$ or $$z_3 = { 2 \over z_1-1} = { 2 \over {1 \over \exp(1/2)-1}-1} = { 2 \exp(1/2)-2 \over 2 - \exp(1/2) }$$ From this $z_5,z_7,...$ follow analoguously, and they are all rational compositions of the "magic" constant $\beta= \frac 1{\sqrt{ e}}$ in the form $${ 1\over z_{2k+1} +1} = a_{2k+1} + b_{2k+1} \beta$$.
For the even indexed $z_{2k}$ the rule is the same, however I did not yet find their "magic constant", say $\gamma$ .