拉马努金连分数证明

拉马努金连分数参考：这里

Here is a famous problem posed by Ramanujan

> Show that $$\left(1 + \frac{1}{1\cdot 3} + \frac{1}{1\cdot 3\cdot 5} + \cdots\right) + \left(\cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}\right) = \sqrt{\frac{\pi e}{2}}$$

The first series seems vaguely familiar if we consider the function $$f(x) = x + \frac{x^{3}}{1\cdot 3} + \frac{x^{5}}{1\cdot 3\cdot 5} + \cdots$$ and note that $$f'(x) = 1 + xf(x)$$ so that $y = f(x)$ satisfies the differential equation $$\frac{dy}{dx} - xy = 1, y(0) = 0$$ The integrating factor here comes to be $e^{-x^{2}/2}$ so that $$ye^{-x^{2}/2} = \int_{0}^{x}e^{-t^{2}/2}\,dt$$ and hence $$f(x) = e^{x^{2}/2}\int_{0}^{x}e^{-t^{2}/2}\,dt$$ Thus the sum of the first series is $$f(1) = \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt$$ But I have no idea about the continued fraction and still more I am not able to figure out how it would simplify to $\sqrt{\pi e/2}$ at the end.

Please provide any hints or suggestions.

**Update**: We have $$\begin{aligned}f(1) &= \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt = \sqrt{e}\int_{0}^{\infty}e^{-t^{2}/2}\,dt - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\\ &= \sqrt{\frac{\pi e}{2}} - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\end{aligned}$$ and hence we finally need to establish $$\sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt = \cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}$$ On further searching in Ramanujan's Collected Papers I found the following formula $$\int_{0}^{a}e^{-x^{2}}\,dx = \frac{\sqrt{\pi}}{2} - \cfrac{e^{-a^{2}}}{2a+}\cfrac{1}{a+}\cfrac{2}{2a+}\cfrac{3}{a+}\cfrac{4}{2a+\cdots}$$ and it seems helpful here. But unfortunately proving this formula seems to be another big challenge for me.

 

This is a sketch of the proof, the details can be found [here][1]. I will offer this sketch because that paper was not intended to prove this result in particular, and I think that a proof might have been written somewhere else.

Consider Mills ratio defined by:
$$\varphi(x)=e^{x^2/2}\int_x^\infty e^{-t^2/2}dt.$$

> **Proposition 1.** There is a *unique* sequence of pairs of polynomials $((P_n,Q_n))_n$ such that $$\varphi^{(n)}(x)=P_n(x)\varphi(x)-Q_n(x)$$
Moreover, these polynomials can be defined inductively by
$$P_{n+1}(x)=xP_n(x)+P'_n,\quad Q_{n+1}=P_n(x)+Q'_n(x)$$
with obvious initial conditions.

The proof in straightforward by induction.

> **Proposition 2.** The sequences $(P_n)_{n}$ and $(Q_n)_{n}$ satisfy the following properties.

> 1. $(P_0,P_1)=(1,x)$, and for all $n\geq1$ we have $P_{n+1}=xP_n+nP_{n-1}$.
> 2. $(Q_0,Q_1)=(0,1)$, and for all $n\geq1$ we have $Q_{n+1}=xQ_n+nQ_{n-1}$.
> 3. For all $n\geq1$ we have $P^\prime_{n}=nP_{n-1}$.

Indeed this follows from Leibniz $n$th derivative formula applied to $\varphi'(x)=x\varphi(x)-1$, and the uniqueness statement in Proposition 1.

> **Proposition 3.** For all $n\geq0$, we have $Q_{n+1}P_n-P_{n+1}Q_n=(-1)^nn!$.

This also an easy induction.

> **Proposition 4.** For all $n\geq0$, $(-1)^n\varphi^{(n)}(x)>0$.

This is a crucial step. Note that
$$\varphi(x)=\int_0^\infty e^{-tx}e^{-t^2/2}dt$$
therefore
$$\varphi^{(n)}(x)=(-1)^n\int_0^\infty t^ne^{-tx}e^{-t^2/2}dt$$

> **Corollary 5.** The sequences $(P_n)_{n }$ and $(Q_n)_{n }$ satisfy the following properties.

> 1. For all $n\geq0$, and all $x>0$, we have
$${Q_{2n}(x)\over P_{2n}(x)}<\varphi(x)<{Q_{2n+1}(x)\over P_{2n+1}(x)}.$$
> 2. For all $n\geq0$, and all $x>0$, we have
$$\left|\varphi(x)-{Q_n(x)\over P_n(x)}\right|<\frac{n!}{P_n(x)P_{n+1}(x)}.$$
> 3. For all $x>0$, we have
$$\lim_{n\to\infty}{Q_n(x)\over P_n(x)}=\varphi(x).$$

This last result, and the recurrence relations from Proposition 2. proves that
$(Q_n/P_n)$ are the convergents of the non regular continued fraction:
$$\frac{Q_{n+1}}{P_{n+1}}=\cfrac{1}{x+\cfrac{1}{x+\cfrac{2}{x+\cfrac{3}{x+\cfrac{4}{x+\cfrac{\ddots}{n/x}}}}}}$$

Finally the desired equality follows from the fact that $\varphi(1)+f(1)=\sqrt{\frac{e\pi}{2}}$, where $f$ is the function considered by the OP.
This concludes the sketch of the proof.$\qquad\square$
[1]: http://arxiv.org/abs/math/0607694

Here is a famous problem posed by Ramanujan

> Show that $$\left(1 + \frac{1}{1\cdot 3} + \frac{1}{1\cdot 3\cdot 5} + \cdots\right) + \left(\cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}\right) = \sqrt{\frac{\pi e}{2}}$$

The first series seems vaguely familiar if we consider the function $$f(x) = x + \frac{x^{3}}{1\cdot 3} + \frac{x^{5}}{1\cdot 3\cdot 5} + \cdots$$ and note that $$f'(x) = 1 + xf(x)$$ so that $y = f(x)$ satisfies the differential equation $$\frac{dy}{dx} - xy = 1, y(0) = 0$$ The integrating factor here comes to be $e^{-x^{2}/2}$ so that $$ye^{-x^{2}/2} = \int_{0}^{x}e^{-t^{2}/2}\,dt$$ and hence $$f(x) = e^{x^{2}/2}\int_{0}^{x}e^{-t^{2}/2}\,dt$$ Thus the sum of the first series is $$f(1) = \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt$$ But I have no idea about the continued fraction and still more I am not able to figure out how it would simplify to $\sqrt{\pi e/2}$ at the end.

Please provide any hints or suggestions.

**Update**: We have $$\begin{aligned}f(1) &= \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt = \sqrt{e}\int_{0}^{\infty}e^{-t^{2}/2}\,dt - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\\ &= \sqrt{\frac{\pi e}{2}} - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\end{aligned}$$ and hence we finally need to establish $$\sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt = \cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}$$ On further searching in Ramanujan's Collected Papers I found the following formula $$\int_{0}^{a}e^{-x^{2}}\,dx = \frac{\sqrt{\pi}}{2} - \cfrac{e^{-a^{2}}}{2a+}\cfrac{1}{a+}\cfrac{2}{2a+}\cfrac{3}{a+}\cfrac{4}{2a+\cdots}$$ and it seems helpful here. But unfortunately proving this formula seems to be another big challenge for me.

The formula given by Ramanujan relating $\pi$ and $e$ is proven in [1] chapter 12 Entry 43 pg.166:
$$\sqrt{\frac{\pi e^x}{2x}}=\frac{1}{x+}\frac{1}{1+}\frac{2}{x+}\frac{3}{1+}\frac{4}{x+}...+\left\{1+\frac{x}{1\cdot3}+\frac{x^2}{1\cdot3\cdot5}+\frac{x^3}{1\cdot3\cdot5\cdot7}+...\right\}$$
The 'hard' term in Ramanujan's formula is the continued fraction. Fortunately the continued fraction can be evaluated in terms of $\textrm{Erfc}(x)$ function. More precicely holds for $Re(b)>0$ ([2] in Appendix pg.578):
$$\lambda(a,b):=\frac{\int^{\infty}_{0}t^a\exp\left(-bt-t^2/2\right)dt}{\int^{\infty}_{0}t^{a-1}\exp\left(-bt-t^2/2\right)dt}=\frac{a}{b+}\frac{a+1}{b+}\frac{a+2}{b+}\frac{a+3}{b+}\dots$$
Set
$$K:=\frac{1}{x+}\frac{1}{1+}\frac{2}{x+}\frac{3}{1+}\frac{4}{x+}...$$
Then one can see easily
$$K=\frac{1}{x+}\frac{\sqrt{x}}{\sqrt{x}+}\frac{2}{\sqrt{x}+}\frac{3}{\sqrt{x}+}\ldots$$
Setting $a=1$ and $b=\sqrt{x}$ in $\lambda(a,b)$, we get
$$K=\frac{1}{x+\sqrt{x}S}$$
where
$$S=\lambda(1,\sqrt{x})=\frac{1}{\sqrt{x}+}\frac{2}{\sqrt{x}+}\frac{3}{\sqrt{x}+}\ldots =\frac{\int^{\infty}_{0}te^{-t\sqrt{x}-t^2/2}}{\int^{\infty}_{0}e^{-t\sqrt{x}-t^2/2}} =\frac{e^{-x/2}\sqrt{\frac{2}{\pi}}}{\textrm{Erfc}\left(\sqrt{\frac{x}{2}}\right)}-\sqrt{x}$$
Hence
$$K=\sqrt{\frac{\pi e^x}{2x}}\textrm{Erfc}\left(\sqrt{\frac{x}{2}}\right)$$
Also the value of the sum in Ramanujan's formula is
$$\sqrt{\frac{e^x\pi}{2}}\textrm{Erf}\left(\sqrt{\frac{x}{2}}\right)$$
From all the above the result follows.

[1]: B.C.Berndt, Ramanujan`s Notebooks Part II. Springer Verlag, New York, 1989.

[2]: L.Lorentzen and H.Waadeland, Continued Fractions with Applications. Elsevier Science Publishers B.V., North Holland, 1992.

About the question for second identity we have:

Let $n$ be non negative integer, then we set
$$G_n(x,y):=\int^{\infty}_{0}t^{x+n}\exp\left(-yt-t^2/2\right)dt$$
With integration by parts we have
$$G_n(a,b)=\int^{\infty}_{0}\frac{d}{dt}\left(\frac{t^{a+n+1}}{a+n+1}\right)\exp\left(-bt-t^2/2\right)dt=$$
$$=0-\int^{\infty}_{0}\frac{t^{a+n+1}}{a+n+1}\exp\left(-bt-t^2/2\right)(-b-t)dt=$$
$$=b\frac{G_{n+1}(a,b)}{a+n+1}+\frac{G_{n+2}(a,b)}{a+n+1}$$
Hence setting $t_n=\frac{G_{n+1}(a,b)}{G_{n}(a,b)}$, $n=0,1,2,\ldots$ we have
$$t_n=\frac{n+a+1}{b+t_{n+1}}$$
and consequently
$$t_0=\frac{G_1(a,b)}{G_0(a,b)}=\lambda(a+1,b)=\frac{a+1}{b+}\frac{a+2}{b+}\frac{a+3}{b+}\ldots$$