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My question concerns the following sequence :

$$u_0=0+\frac{1}{2} \quad u_1 =0+\frac{1}{2+\frac{3}{4}} = \frac{4}{11} \quad u_2=0+\frac{1}{2+\frac{3}{4+\frac{5}{6}}}=\frac{29}{76} \quad u_3=0+\frac{1}{2+\frac{3}{4+\frac{5}{6+\frac{7}{8}}}}=\frac{52}{137}$$

It is distinct from the harmonic continued fraction : Convergence of a Harmonic Continued Fraction

It can be defined by the recurrence relation : $$v_n=2n+1-\frac{1}{2n+2} \qquad v_{k}=2k+\frac{2k+1}{v_{k+1}} \quad (0\leq k <n) (*)$$

Applying the formula (*) $n$ times, we have $u_n=v_0$.

A simple Python script gives the following results : $$ \frac{1}{2}, \frac{4}{11}, \frac{29}{76}, \frac{52}{137}, \frac{2861}{7534}, \frac{37192}{97943}, \frac{557881}{1469144} $$ The sequence seems to converge to $\ell\approx 0.37973195474099564$

from fractions import Fraction
def f(n):
    v = Fraction(2 * n + 1, 1) - Fraction(1, 2 * n + 2)
    for i in range(n - 1, -1, -1):
        v = Fraction(2 * i, 1) + Fraction(2 * i + 1, v)
    return v

print([f(k) for k in range(10)])

Here is my question : what is the exact value of $\ell$ and how can we prove it?

Tito Piezas III
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alati ahmad
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  • Your continued fraction, $$0+\cfrac { 1 }{ 2+\cfrac { 3 }{ 4+\cfrac { 5 }{ 6+\cfrac { 7 }{ 8+\ddots } } } } =,??\qquad$$ has a complementary cfrac with a closed-form, $$1+\cfrac { 2 }{ 3+\cfrac { 4 }{ 5+\cfrac { 6 }{ 7+\cfrac { 8 }{ 11+\ddots } } } } =\frac{1}{\sqrt{e}-1}$$ Thus, whether yours has a closed-form too is interesting to consider. See also https://mathworld.wolfram.com/ContinuedFractionConstants.html. – Tito Piezas III Dec 25 '23 at 09:46
  • The 2nd cfrac above has a typo. It should be $1,3,5,7,9,$ etc. A 3rd cfrac with all odd terms also has a closed-form $$\qquad 1+\cfrac{1}{3+\cfrac{3}{5+\cfrac{5}{7+\cfrac{7}{9+\ddots}}}}= \frac{I_0(1/4)+I_1(1/4)}{I_0(1/4)-I_1(1/4)}$$ See this post. It is the 4th cfrac with all even terms that doesn't seem to have a closed-form. – Tito Piezas III Dec 29 '23 at 14:01

2 Answers2

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According to sequence A113014 of the OEIS, $$\ell = \frac{\sqrt{2e/\pi}}{\operatorname{erfi}{(1/\sqrt{2})}} - 1, $$

where $\operatorname{erfi}(\cdot)$ is the imaginary error function.

I don't know how to prove this, yet.

Max Muller
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    +1 Based on a recent post of mine, I had a feeling either $e$, or $\pi$, or $\operatorname{erf}(x)$ would pop up. Never knew $\operatorname{erf}$ had an imaginary version. – Tito Piezas III Dec 25 '23 at 10:22
  • @OscarLanzi Ramanujan's nice series+cfrac identity for $\sqrt{\dfrac{\pi,e}{2}}$ can be expressed as two cfracs, \begin{align}\sqrt{\frac{\pi,e}{2}} &=1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}} ; + ; \cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}\ &=\quad\sqrt{\frac{\pi,e}{2}} \operatorname{erf}\Big(\sqrt{\tfrac 1 2}\Big)\qquad+\qquad\sqrt{\frac{\pi,e}{2}} \left(1-\operatorname{erf}\Big(\sqrt{\tfrac 1 2}\Big)\right)\end{align} so makes me wonder if $\ell$ can be split into a series+cfrac as well. – Tito Piezas III Dec 25 '23 at 10:50
  • @OscarLanzi Thanks for the info, though a minor correction since Mathworld says $\operatorname{erfi}(x) = -i\operatorname{erf}(i,x)$. – Tito Piezas III Dec 25 '23 at 13:47
  • @TitoPiezasIII we simply have $ i\operatorname{erfi}(x)=\operatorname{erf}(ix)=(2/\sqrt\pi)\int_0^x\exp(t^2)dt.$ – Oscar Lanzi Dec 25 '23 at 14:08
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    Comments are now out of order, but I added the factor of $i$. We really need to have comments indefinitely editable by their owners. I remember asking for this on Meta and basically getting blown off. – Oscar Lanzi Dec 25 '23 at 14:11
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This is an addendum to Muller's answer. Since he found out that the OP's cfrac $\ell$ involved the error function $\operatorname{erf}(x)$, we can connect it to Ramanujan's beautiful identity,

$$\sqrt{\frac{\pi\,e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\,\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$

The closed-form of its infinite series part is,

$$\sum_{n=0}^\infty \prod_{k=0}^n\frac{x^n}{2k+1} = \sum_{n=0}^\infty \frac{x^n}{(2n+1)!!} = \sqrt{\frac{\pi\,e^x}{2x}} \operatorname{erf}\Big(\sqrt{\tfrac x 2}\Big)$$

Therefore at $x=-1$ we have the imaginary error function,

\begin{align}\sqrt{\frac{\pi}{2\,e}} \operatorname{erfi}\Big(\sqrt{\tfrac 1 2}\Big) &= 1-\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}-\frac{1}{1\cdot3\cdot5\cdot7}+\dots\\ &= \frac{1}{\color{red}1+\ell} = \cfrac1{\color{red}1+\cfrac{1}{2+\cfrac{3}{4+\cfrac{5}{6+\ddots}}}}\\ &=0.724778459007076331\dots \end{align}

which is OEIS A306858.

Tito Piezas III
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  • Nice explanation (though the mentioned closed form seems to be difficult to prove https://math.stackexchange.com/questions/833920/an-infinite-series-plus-a-continued-fraction-by-ramanujan). Thanks! – alati ahmad Dec 26 '23 at 19:17
  • @alatiahmad At least now you have a closed-form for your continued fraction in terms of the error function, or express it as a nice alternating series (the non-alternating version being used by Ramanujan). It was a productive post. – Tito Piezas III Dec 27 '23 at 03:10