Does anyone know the exact value for the continued fraction of $$1+\cfrac{1}{3+\cfrac{3}{5+\cfrac{5}{7+\cfrac{7}{9+\ddots}}}}?$$

I already know that $$1+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{7+\cfrac{1}{9\ddots}}}}=\frac{e^2+1}{e^2-1},$$ but I only figured that out by typing the decimal approximation into google of the first few terms of the continued fraction (before I knew the exact value) which took me to a math paper saying that $\frac{e^2+1}{e^2-1}$ roughly equals the decimal approximation I typed in. I then typed in the continued fraction of $\frac{e^2+1}{e^2-1}$ into wolfram alpha and it spat out $$1+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{7+\cfrac{1}{9\ddots}}}}.$$ I have no idea how to solve these so please don't downvote, I'm just doing this in case it's useful to someone one day, and out of curiosity of course.

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    $\begingroup$ @MaximilianJanisch Note that the question you linked to is already mentioned by the OP. $\endgroup$
    – Trebor
    Commented Jun 30, 2021 at 15:53
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    $\begingroup$ Note we can rewrite as $1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\cfrac{1/9}{1+\ddots}}}}$ so that $(a_{2k+1}-1)a_{2k+3}=1/(2k+1)$ for each $k\ge0$. $\endgroup$
    – TheSimpliFire
    Commented Jun 30, 2021 at 16:10
  • $\begingroup$ FWIW, the ISC did not return any results. $\endgroup$ Commented Jun 30, 2021 at 16:14
  • $\begingroup$ But Śleszyński–Pringsheim does say that your CF converges. $\endgroup$ Commented Jun 30, 2021 at 16:25

1 Answer 1


Comments point out that this post did the wrong continued fraction. For the correct one, use $a=-1$ not $1$. Then follow the same Satz $2$. The result is $$ \frac{2\;{}_2F_1(-\frac12;1;\frac12)}{{}_1F_1(\frac12;2;\frac12)} =\frac{I_0(\frac14)+I_1(\frac14)}{I_0(\frac14)-I_1(\frac14)} \approx 1.2831923 . $$

original post
Here is the reference for everything on continued fractions (as of 1913):

Perron, Oskar, Die Lehre von den Kettenbrüchen., Leipzig - Berlin: B. G. Teubner. xiii, 520 S. $8^\circ$ (1913). ZBL43.0283.04.

Section 81, Satz 2 evaluates

$$ c + \frac{a+b}{\displaystyle c+d + \frac{a+2b}{\displaystyle c+2d+\frac{a+3b}{\displaystyle c+3d+\ddots}}} $$

So we need $a=1,b=2,c=1,d=2$.

Value of the continued fraction is $$ \frac{2\;{}_1F_1(\frac12, 1, \frac12)}{{}_1F_1(\frac32, 2, \frac12)} \approx 1.779306397 $$ This can be written $$ \frac{2e^{1/4} I_0(\frac14)} {e^{1/4} I_0(\frac14)+e^{1/4} I_1(\frac14)} = \frac{2}{\displaystyle 1+\frac{I_1(\frac14)}{I_0(\frac14)}} $$ in terms of Bessel functions.

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    $\begingroup$ I am far away from my copy of Perron's book to verify, but your proposed value of $1.779306397$ is fairly different from the $1.283192342$ one might obtain from numerically evaluating OP's CF. $\endgroup$ Commented Jun 30, 2021 at 22:35
  • $\begingroup$ Checking with the CF of ${}_1 F_1$ that I am familiar with, what you evaluated was in fact $$1+\cfrac{3}{3+\cfrac{5}{5+\cfrac{7}{7+\cfrac{9}{9+\cdots}}}}$$ which is a completely different thing altogether. $\endgroup$ Commented Jun 30, 2021 at 22:59
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    $\begingroup$ To get the OP's CF, you need to set $a=-1$, not $a=1$. $\endgroup$ Commented Jun 30, 2021 at 23:11
  • $\begingroup$ Using @Barry's correction, the expected closed form becomes $$\frac{2\,{}_1 F_1\left({{-1/2}\atop{1}}\middle|1/2\right)}{{}_1 F_1\left({{1/2}\atop{2}}\middle|1/2\right)}=\cfrac2{1-\cfrac{I_1\left(\frac14\right)}{I_0\left(\frac14\right)}}-1$$ which now tracks with the numeric evaluation. $\endgroup$ Commented Jun 30, 2021 at 23:35
  • $\begingroup$ @J.M.ain'tamathematician That is, indeed, the CF I compute here. Thanks for the correction. $\endgroup$
    – GEdgar
    Commented Jul 1, 2021 at 0:09

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