Find an explicit real number that is not in the set produced by the convergent rational series

Question

Let $$\mathcal L=\left\{\sum_{n=1}^{\infty}a_n:\exists P(x),Q(x)\in\mathbb{Z}[x],\forall k\in\mathbb{N},Q(k)\neq 0,a_n=\frac{P(n)}{Q(n)},\sum_{n=1}^{\infty}a_n\text{ converges}\right\}.$$ We have $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=1$, hence for all $a/b\in\mathbb{Q}$ we have $a/b\in\mathcal L$, so that $\mathbb{Q}\subseteq\mathcal L$. Furthermore, $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, thus $\mathcal L\neq\mathbb{Q}$. In addition, $\mathcal L$ is countable, and therefore $\mathcal L\neq \mathbb{R}$. This is all I can say about $\mathcal L$ for now. A natural question arises.

Do we know of an explicit real number that is not in $\mathcal L$ (from what we showed such a number exists and is irrational)?

Answer 1

Every value of $\mathcal L$ is computable, hence any fixed noncomputable real such as a Chaitin halting probability $\Omega_U$ does not lie in $\mathcal L$. Indeed, let $a_n=P(n)/Q(n)$ with $P,Q\in\mathbb Z[x]$, $\deg P=d$, $\deg Q=D$, and $Q(k)\neq0$ for all $k$. If $D-d\le1$ then $$a_n=\frac{p_d}{q_D}\,\frac1{n^{D-d}}+O\!\left(\frac1{n^{D-d+1}}\right),$$ so $a_n\sim c/n$ when $D-d=1$ or $a_n\to c\neq0$ when $D-d\le0$. Thus $\sum a_n$ cannot converge, and convergence forces $D-d\ge2$.

Write $B=\sum_i |p_i|$ for the coefficients of $P$, let $q_D$ be the leading coefficient of $Q$, and set $S=\sum_{i<D}|q_i|$. Take $$N=\max\!\{2,\;1+\lceil \tfrac{2S}{|q_D|}\rceil\}.$$ For $n\ge N$ we have $|P(n)|\le B n^d$ and $|Q(n)|\ge |q_D|n^D - S n^{D-1}\ge \tfrac{|q_D|}{2}n^D$, hence $$|a_n|\le \frac{2B}{|q_D|}\,n^{-(D-d)}\le \frac{K}{n^2}\qquad (K:=2B/|q_D|).$$ Therefore $\sum_{n\ge M}|a_n|\le K/(M-1)$ for every $M\ge N$. Given $\varepsilon>0$, choose $$M\ge \max\!\{N,\;1+\lceil K/\varepsilon\rceil\}$$ to get $\sum_{n\ge M}|a_n|\le \varepsilon$. This produces a computable modulus of Cauchy convergence, so $\sum_{n=1}^\infty a_n$ is a computable real. Since $\Omega_U$ is noncomputable, it follows that $\Omega_U\notin\mathcal L$.

Moreover, there is an explicit computable real outside $\mathcal L$: effectively enumerate all pairs $(P_i,Q_i)$ with $Q_i(k)\neq0$ and $\deg Q_i\ge\deg P_i+2$, let $s_i=\sum_{n\ge1}P_i(n)/Q_i(n)$ (each $s_i$ computable by the bound above), write $s_i$ in canonical binary expansion and let $d_i$ be its $2i$-th binary digit. Define $\xi=0.b_1b_2b_3\ldots$ by setting $b_{2i-1}:=0$ and $b_{2i}:=1-d_i$. Then $\xi\notin\mathcal L$ and $\xi$ is computable.