¿Cuál es el valor de $\frac 1 {0! + 1! + 2!} +\frac 1 {1! + 2! + 3!} + \frac 1 { 2! + 3! + 4!} + …?$

Question

¿Cuál es el valor de x?

$$e = 1 / 0! + 1 / 1! + 1 / 2! + … .$$

$$1 = 1 / (0! + 1!) + 1 / (1! + 2!) + 1 / (2! + 3!) + … .$$

$$(x = ) 1 / (0! + 1! + 2!) + 1 / (1! + 2! + 3!) + 1 / (2! + 3! + 4!) + … .$$

En mi cálculo, mediante la programación, x se acerca 0.40037967700464134050027862710343065978234584790717558212650643072643052259740811195942853169077421.

Si es posible encontrar el valor de x, por favor, dime el valor.

Answer 1

Tenga en cuenta que $$k! + (k+1)! + (k+2)! = k!(k+2)^2,$$ so that the sum $$S = \sum_{k=0}^\infty \frac{1}{k!+(k+1)!+(k+2)!} = \sum_{k=0}^\infty \frac{1}{(k+2)^2 k!}.$$ This suggests considering the function $$f(z) = z e^z = \sum_{k=0}^\infty \frac{z^{k+1}}{k!}.$$ Taking the integral gives $$g(z) = \int_{t=0}^z f(t) \, dt = (z-1)e^z + 1 = \sum_{k=0}^\infty \frac{z^{k+2}}{(k+2)k!}.$$ Dividing by $z$ and integrating once again, we get $$\sum_{k=0}^\infty \frac{z^{k+2}}{(k+2)^2 k!} = \int_{t=0}^z \frac{g(t)}{t} \, dt = e^z - 1 + \int_{t=0}^z \frac{e^t - 1}{t} \, dt,$$ for which the choice $z = 1$ yields $$S = e - 1 + \int_{t=0}^1 \frac{e^t-1}{t} \, dt.$$ This last integral doesn't have an elementary closed form; Mathematica evaluates it as $$\gamma - \operatorname{Ei}(1),$$ where $$\operatorname{Ei}(z) = -\int_{t=-z}^\infty \frac{e^{-t}}{t} \, dt.$$