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The following identity is due to Ramanujan:

$$\DeclareMathOperator{\k}{\vphantom{\sum}\vcenter{\LARGE K}} \sqrt{\frac{\pi e}{2}}=\frac{1}{1+\k_{n=1}^\infty \frac{n}{1}}+\sum_{n=0}^\infty\frac{1}{(2n+1)!!}$$ or $$\sqrt{\frac{\pi e}{2}}=\cfrac{1}{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{\ddots}}}}+1+\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\cdots $$

I'm interested in the proof of this identity, but I couldn't manage to find any reference except for the linked page.

metamorphy
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sting890
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1 Answers1

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As pointed out by the linked page, it suffices to prove

$$ 1+\dfrac{1}{1+\dfrac{2}{1+\dfrac{3}{1+\ddots}}} = \sqrt{\frac{2}{\pi e}} \frac{1}{\operatorname{erfc}(1/\sqrt{2})}. \tag{1} $$

To this end, we will resort to the standard theory of continued fraction. Define $(p_n)$ and $(q_n)$ by the following relation:

$$ \begin{pmatrix} p_n \\ q_n \end{pmatrix} = A_1A_2\dots A_n \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad\text{where}\quad A_n = \begin{pmatrix} 1 & n \\ 1 & 0 \end{pmatrix}. $$

Then it is routine to check that

\begin{align*} p_0 &= 1, & p_1 &= 1, & p_{n+2} &= p_{n+1} + (n+1) p_n, \\ q_0 &= 0, & q_1 &= 1, & q_{n+2} &= q_{n+1} + (n+1) q_n. \end{align*}

Moreover, if $f_A(z) = \frac{a_{11}z+a_{12}}{a_{21}z+a_{22}}$ denotes the linear fractional transformation induced by the $2\times2$ matrix $A=[a_{ij}]_{1\leq i,j\leq 2}$, then we have:

$$ \frac{p_n}{q_n} = f_{A_1\dots A_n}(\infty) = (f_{A_1}\circ\dots\circ f_{A_n})(\infty) = 1+\dfrac{1}{1+\dfrac{2}{\ddots+\dfrac{\ddots}{1+\dfrac{n-1}{1}}}} $$

The standard theory also affirms that this converges as $n\to\infty$. So it suffices to compute the limit as $n\to\infty$. To this end, note that both $p_n$ and $q_n$ are increasing and diverges to $\infty$. Moreover, if we introduce the exponential generating functions of $(p_n)$ and $(q_n)$ by

$$ y_p (x) = \sum_{n=0}^{\infty} \frac{p_n}{n!}x^n \quad\text{and}\quad y_q (x) = \sum_{n=0}^{\infty} \frac{q_n}{n!}x^n, $$

then they satisfy

$$ y_p' = (1+x)y_p \quad\text{and}\quad y_q' = 1 + (1+x)y_q. $$

These equations, together with the initial conditions $y_p(0) = p_0 = 1$ and $y_q(0) = q_0 = 0$, can be solved by the integrating factor method, and we obtain

$$ y_p(x) = e^{x+\frac{x^2}{2}} \quad \text{and} \quad y_q(x) = e^{x+\frac{x^2}{2}}\sqrt{\frac{\pi e}{2}} \left( \operatorname{erf}\left(\frac{1+x}{\sqrt{2}}\right) - \operatorname{erf}\left(\frac{1}{\sqrt{2}}\right) \right). $$

Now by invoking the standard argument for abelian theorem,

$$ \lim_{n\to\infty} \frac{p_n}{q_n} = \lim_{x\to\infty} \frac{y_p(x)}{y_q(x)} = \sqrt{\frac{2}{\pi e}} \frac{1}{\operatorname{erfc}\left(1/\sqrt{2}\right)} $$

as required.

Sangchul Lee
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  • That's a very cool technique! +1 Can you provide some reference where this is discussed in detail? – Paramanand Singh Aug 23 '20 at 03:13
  • @ParamanandSingh, Glad you enjoyed my answer! The first half comes from the standard theory of generalized continued fractions, and I learned it from a couple of online lecture notes. The remaining half using ODE of EGF is self-concocted, although the answer in one of my old questions (see [this](https://math.stackexchange.com/a/3751976/9340)) tells that it is in fact a well-known approach. – Sangchul Lee Aug 25 '20 at 19:28