# The image set of the operator $\sum_{n=2}^\infty a_n\frac{\Gamma(n)}{n^{n-1}}$, for sequences with $a_1=0$ and $a_n\in\{-1,0,1\},\forall n\geq 2$

For a sequence $(a_n)_{n\geq 1}$ with $a_1=0$ and $a_k\in\{-1,0,1\}$ for all $k\geq 2$ one can to prove the convergence of $$\sum_{n=2}^\infty a_n\frac{\Gamma(n)}{n^{n-1}},\tag{1}$$ where $\Gamma(s)$ is the Gamma function, for example using [1] $$\left| \sum_{n=2}^\infty a_n\frac{\Gamma(n)}{n^{n-1}} \right|\ \leq 1.$$

Definition. On the set of our sequences $$\mathcal{S}= \left\{a= (a_n)_{n\geq 1}: a_1=0, \forall k\geq 2 \text{ with }a_k\in\{-1,0,1\}\right\} \tag{2}$$ we define $\mathcal{F}:\mathcal{S}\to [-\xi,\xi]$, where $\xi=\sum_{n=2}^\infty 1\cdot\frac{\Gamma(n)}{n^{n-1}}$ to be $$\mathcal{F}(a)=\sum_{n=2}^\infty a_n\frac{\Gamma(n)}{n^{n-1}}.\tag{3}$$

Question. Is $\mathcal{F}(\mathcal{S})=[-\xi,\xi]$? In other words, my question is if we can to prove that our operator is surjective:

$\qquad$For each fixed $x^{*}\in[-\xi,\xi]$ there exists a sequence $a^{*}\in\mathcal{S}$ such that $\mathcal{F}(a^*)=x^*$.

Many thanks.

This is a table containing examples of an approximation of $\mathcal{F}(a)$ for the corresponding sequences for $n\geq 2$ defined as $a_n=1$, $a_n=n\text{ mod } 2$, $a_n=(-1)^{n-1}$, $a_n=\mu(n)$ the Möbius function and $\lambda(n)$ the Liouville function.

$$\begin{array}{ll} \hfill a_n, n\geq 2\hfill & \hfill\sum_{n=2}^{50}a_n\frac{\Gamma(n)}{n^{n-1}}\hfill \\ \hline \\ \qquad 1 & \quad \approx 0.87985\\ \\ \qquad n\text{ mod} 2 & \quad \approx 0.26784\\ \\ \qquad (-1)^{n-1} & \quad \approx -0.34417\\ \\ \qquad \mu(n) & \quad \approx -0.75109\\ \\ \qquad \lambda(n) & \quad \approx -0.65887 \end{array}$$

## References:

[1] Juan-Bosco Romero Márquez, Problem 1651, Mathematics Magazine, Vol. 75, No. 3 (June, 2002).

• I've deleted previous comment since now is without context, but since previous comment included a congratulations message I add here it one more time: congratulations for the diamond @JackD'Aurizio – user243301 Nov 12 '17 at 17:58

$\Gamma(n)/n^{n-1} =n!/n^n \approx \sqrt{2\pi n}/e^n$.
• This is relevant for sure. Now, given that $2<e<3$, is the operator $\mathcal{F}$ surjective or $\mathcal{F}(\mathcal{S})$ is a Cantor set with measure zero? – Jack D'Aurizio Oct 23 '17 at 15:45
• Almost looks like the representation of numbers in base $e$. – marty cohen Oct 23 '17 at 15:48
• Many thanks, was a curiosity when I was calculating with these series of cited form (also was doing calculations to get the asymptotic behaviour of different series as $\sum_{n=2}^N (p_n\text{ mod }4)\frac{\Gamma(n)}{n^{n-1}}$, where $p_n$ is the $n$th prime number), when I wondered what about the image of my operator. – user243301 Oct 23 '17 at 15:51