Aquí es un famoso problema planteado por Ramanujan
Mostrar que $$\left(1 + \frac{1}{1\cdot 3} + \frac{1}{1\cdot 3\cdot 5} + \cdots\right) + \left(\cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}\right) = \sqrt{\frac{\pi e}{2}}$$
La primera serie parece vagamente familiar si consideramos que la función $$f(x) = x + \frac{x^{3}}{1\cdot 3} + \frac{x^{5}}{1\cdot 3\cdot 5} + \cdots$$ 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 $\sqrt{\pi e/2}$ al final.
Por favor proporcione consejos o sugerencias.
Actualización: Tenemos $$\begin{aligned}f(1) &= \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt = \sqrt{e}\int_{0}^{\infty}e^{-t^{2}/2}\,dt - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\\ &= \sqrt{\frac{\pi e}{2}} - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\end{aligned}$$ 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 $$\int_{0}^{a}e^{-x^{2}}\,dx = \frac{\sqrt{\pi}}{2} - \cfrac{e^{-a^{2}}}{2a+}\cfrac{1}{a+}\cfrac{2}{2a+}\cfrac{3}{a+}\cfrac{4}{2a+\cdots}$$ y es que parece ser útil aquí. Pero, por desgracia probar esta fórmula parece ser un gran desafío para mí.
