An infinite series plus a continued fraction by Ramanujan
Solution 1:
This is a sketch of the proof, the details can be found here. 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.
- $(P_0,P_1)=(1,x)$, and for all $n\geq1$ we have $P_{n+1}=xP_n+nP_{n-1}$.
- $(Q_0,Q_1)=(0,1)$, and for all $n\geq1$ we have $Q_{n+1}=xQ_n+nQ_{n-1}$.
- 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.
- For all $n\geq0$, and all $x>0$, we have $${Q_{2n}(x)\over P_{2n}(x)}<\varphi(x)<{Q_{2n+1}(x)\over P_{2n+1}(x)}.$$
- For all $n\geq0$, and all $x>0$, we have $$\left|\varphi(x)-{Q_n(x)\over P_n(x)}\right|<\frac{n!}{P_n(x)P_{n+1}(x)}.$$
- 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.$\qquad\square$
Solution 2:
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 $$ t_0=\frac{G_1(a,b)}{G_0(a,b)}=\lambda(a+1,b)=\frac{a+1}{b+}\frac{a+2}{b+}\frac{a+3}{b+}\ldots $$