拉马努金连分数参考：这里
Here is a famous problem posed by Ramanujan
> Show that $$left(1 + frac{1}{1cdot 3} + frac{1}{1cdot 3cdot 5} + cdots ight) + left(cfrac{1}{1+}cfrac{1}{1+}cfrac{2}{1+}cfrac{3}{1+}cfrac{4}{1+cdots} ight) = sqrt{frac{pi e}{2}}$$
The first series seems vaguely familiar if we consider the function $$f(x) = x + frac{x^{3}}{1cdot 3} + frac{x^{5}}{1cdot 3cdot 5} + cdots$$ 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 $$egin{aligned}f(1) &= sqrt{e}int_{0}^{1}e^{-t^{2}/2}\,dt = sqrt{e}int_{0}^{infty}e^{-t^{2}/2}\,dt - sqrt{e}int_{1}^{infty}e^{-t^{2}/2}\,dt\
&= sqrt{frac{pi e}{2}} - sqrt{e}int_{1}^{infty}e^{-t^{2}/2}\,dtend{aligned}$$ 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 $ngeq1$ we have $P_{n+1}=xP_n+nP_{n-1}$.
> 2. $(Q_0,Q_1)=(0,1)$, and for all $ngeq1$ we have $Q_{n+1}=xQ_n+nQ_{n-1}$.
> 3. For all $ngeq1$ we have $P^prime_{n}=nP_{n-1}$.
Indeed this follows from Leibniz $n$th derivative formula applied to $varphi'(x)=xvarphi(x)-1$, and the uniqueness statement in Proposition 1.
> **Proposition 3.** For all $ngeq0$, we have $Q_{n+1}P_n-P_{n+1}Q_n=(-1)^nn!$.
This also an easy induction.
> **Proposition 4.** For all $ngeq0$, $(-1)^nvarphi^{(n)}(x)>0$.
This is a crucial step. Note that
$$varphi(x)=int_0^infty e^{-tx}e^{-t^2/2}dt$$
therefore
$$varphi^{(n)}(x)=(-1)^nint_0^infty t^ne^{-tx}e^{-t^2/2}dt$$
> **Corollary 5.** The sequences $(P_n)_{n }$ and $(Q_n)_{n }$ satisfy the following properties.
> 1. For all $ngeq0$, 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 $ngeq0$, and all $x>0$, we have
$$left|varphi(x)-{Q_n(x)over P_n(x)}
ight|<frac{n!}{P_n(x)P_{n+1}(x)}.$$
> 3. For all $x>0$, we have
$$ lim_{n oinfty}{Q_n(x)over P_n(x)}=varphi(x).$$
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 $varphi(1)+f(1)=sqrt{frac{epi}{2}}$, where $f$ is the function considered by the OP.
This concludes the sketch of the proof.$qquadsquare$
[1]: http://arxiv.org/abs/math/0607694
Here is a famous problem posed by Ramanujan
> Show that $$left(1 + frac{1}{1cdot 3} + frac{1}{1cdot 3cdot 5} + cdots ight) + left(cfrac{1}{1+}cfrac{1}{1+}cfrac{2}{1+}cfrac{3}{1+}cfrac{4}{1+cdots} ight) = sqrt{frac{pi e}{2}}$$
The first series seems vaguely familiar if we consider the function $$f(x) = x + frac{x^{3}}{1cdot 3} + frac{x^{5}}{1cdot 3cdot 5} + cdots$$ 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 $$egin{aligned}f(1) &= sqrt{e}int_{0}^{1}e^{-t^{2}/2}\,dt = sqrt{e}int_{0}^{infty}e^{-t^{2}/2}\,dt - sqrt{e}int_{1}^{infty}e^{-t^{2}/2}\,dt\
&= sqrt{frac{pi e}{2}} - sqrt{e}int_{1}^{infty}e^{-t^{2}/2}\,dtend{aligned}$$ 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:
$$
sqrt{frac{pi e^x}{2x}}=frac{1}{x+}frac{1}{1+}frac{2}{x+}frac{3}{1+}frac{4}{x+}...+left{1+frac{x}{1cdot3}+frac{x^2}{1cdot3cdot5}+frac{x^3}{1cdot3cdot5cdot7}+...
ight}
$$
The 'hard' term in Ramanujan's formula is the continued fraction. Fortunately the continued fraction can be evaluated in terms of $ extrm{Erfc}(x)$ function. More precicely holds for $Re(b)>0$ ([2] in Appendix pg.578):
$$
lambda(a,b):=frac{int^{infty}_{0}t^aexpleft(-bt-t^2/2
ight)dt}{int^{infty}_{0}t^{a-1}expleft(-bt-t^2/2
ight)dt}=frac{a}{b+}frac{a+1}{b+}frac{a+2}{b+}frac{a+3}{b+}dots
$$
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
$$
S=lambda(1,sqrt{x})=frac{1}{sqrt{x}+}frac{2}{sqrt{x}+}frac{3}{sqrt{x}+}ldots
=frac{int^{infty}_{0}te^{-tsqrt{x}-t^2/2}}{int^{infty}_{0}e^{-tsqrt{x}-t^2/2}}
=frac{e^{-x/2}sqrt{frac{2}{pi}}}{ extrm{Erfc}left(sqrt{frac{x}{2}}
ight)}-sqrt{x}
$$
Hence
$$
K=sqrt{frac{pi e^x}{2x}} extrm{Erfc}left(sqrt{frac{x}{2}}
ight)
$$
Also the value of the sum in Ramanujan's formula is
$$
sqrt{frac{e^xpi}{2}} extrm{Erf}left(sqrt{frac{x}{2}}
ight)
$$
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
$$
G_n(x,y):=int^{infty}_{0}t^{x+n}expleft(-yt-t^2/2
ight)dt
$$
With integration by parts we have
$$
G_n(a,b)=int^{infty}_{0}frac{d}{dt}left(frac{t^{a+n+1}}{a+n+1}
ight)expleft(-bt-t^2/2
ight)dt=
$$
$$
=0-int^{infty}_{0}frac{t^{a+n+1}}{a+n+1}expleft(-bt-t^2/2
ight)(-b-t)dt=
$$
$$
=bfrac{G_{n+1}(a,b)}{a+n+1}+frac{G_{n+2}(a,b)}{a+n+1}
$$
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
$$
t_0=frac{G_1(a,b)}{G_0(a,b)}=lambda(a+1,b)=frac{a+1}{b+}frac{a+2}{b+}frac{a+3}{b+}ldots
$$