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$$\int_{1}^{\infty}\frac{((\left \lfloor x \right \rfloor-1)!)^2}{(2\left \lfloor x \right \rfloor-1)!}dx$$

What I do so far

$$\int_{1}^{\infty}\frac{((\left \lfloor x \right \rfloor-1)!)^2}{(2\left \lfloor x \right \rfloor-1)!}dx = \sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{((n-1)!)^2}{(2n-1)!}dx$$

$$\int_{n}^{n+1}\frac{((n-1)!)^2}{(2n-1)!}dx = \frac{((n-1)!)^2}{(2n-1)!} \cdot (n+1 - n) = \frac{((n-1)!)^2}{(2n-1)!}$$

$$\sum_{n=1}^{\infty}\frac{((n-1)!)^2}{(2n-1)!}$$ = \begin{align*} \frac{2\pi}{3\sqrt{3}} \end{align*}

Martin.s
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  • Yes, the area under the graph of the integrand is just rectangles with width one, so it's the same as summing the heights of those rectangles. I might look at the first few terms of the sum for a nice pattern (but if the answer involves $\pi$ like that, it maybe won't be so simple). Otherwise I'd try finding a formula for the generating function. – aschepler Oct 04 '23 at 23:52
  • How did you compute your last sum of the factorials? – Benjamin Wang Oct 05 '23 at 00:08
  • Wolfram alpha? Amy ideas how to solve last sum? – Martin.s Oct 05 '23 at 00:16

1 Answers1

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Your sum can be evaluated from a special case of the more general series expansion $$\arcsin^2 x = \frac{1}{2}\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2 \binom{2n}{n}}. \tag{1}$$ Taking the derivative, we obtain $$\begin{align} \frac{2 \arcsin x}{\sqrt{1-x^2}} &= 2\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{n \binom{2n}{n}} \\ &= 2 \sum_{n=1}^\infty \frac{n! \, (n-1)!}{(2n)!}(2x)^{2n-1} \\ &= 2 \sum_{n=1}^\infty \frac{n}{2n} \frac{(n-1)!^2}{(2n-1)!} (2x)^{2n-1} \\ &= \sum_{n=1}^\infty \frac{(n-1)!^2}{(2n-1)!} (2x)^{2n-1}. \tag{2} \end{align}$$ Your sum is $(2)$ evaluated at $x = 1/2$: $$\sum_{n=1}^\infty \frac{(n-1)!^2}{(2n-1)!} = \frac{2 \arcsin \frac{1}{2}}{\sqrt{3/4}} = \frac{2\pi}{3\sqrt{3}}. \tag{3}$$ The identity $(1)$ is Equation $(19)$ in the MathWorld entry for inverse sine, which cites Borwein (2004) but as I do not have this text, I cannot provide a proof at this time.

heropup
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