Infinite series with floor function
- Tolaso J Kos
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Infinite series with floor function
Let $\lfloor x \rfloor$ denote the floor function. Prove that:
$$\sum \limits_{n=0}^{\infty} \frac{1} {2^n\cdot n!}\left \lfloor \frac{n!}{e} \right \rfloor= \frac{2}{\sqrt{e}}-1$$
$$\sum \limits_{n=0}^{\infty} \frac{1} {2^n\cdot n!}\left \lfloor \frac{n!}{e} \right \rfloor= \frac{2}{\sqrt{e}}-1$$
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Re: Infinite series with floor function
It seems that the value of the series is $\frac{3-e}{2\sqrt{e}}$. To this end, recall the exponentinal generating function of $\lfloor n!/e\rfloor$ (due to Mark Hoeij, 2011)
$$\sum_{n=0}^\infty\,\frac{1}{n!}\,\left\lfloor\frac{n!}{e}\right\rfloor\,x^n = -\frac{1}{2}\left(e^x + \frac{x+1}{x-1}\,e^{-x}\right).$$
Setting $x = 1/2$ yields
$$\sum_{n=0}^\infty\,\frac{1}{2^n\,n!}\,\left\lfloor\frac{n!}{e}\right\rfloor = \frac{3-e}{2\sqrt{e}}$$
as claimed.
$$\sum_{n=0}^\infty\,\frac{1}{n!}\,\left\lfloor\frac{n!}{e}\right\rfloor\,x^n = -\frac{1}{2}\left(e^x + \frac{x+1}{x-1}\,e^{-x}\right).$$
Setting $x = 1/2$ yields
$$\sum_{n=0}^\infty\,\frac{1}{2^n\,n!}\,\left\lfloor\frac{n!}{e}\right\rfloor = \frac{3-e}{2\sqrt{e}}$$
as claimed.
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