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\begin{document}

\begin{theorem}\label{maintheorem}
There exists a sequence $(b_k)_{k=1}^{\infty}$ with values in the set $\{1,2,3,4,5\}$ such that the infinite sum $\sum_{k=1}^{\infty} \frac{1}{2^k+b_k}$ is a rational number.
\end{theorem}

%?? There exists a bounded sequence of non-zero integers $(b_k)_{k=1}^{\infty}$ such that the infinite sum $\sum_{k=1}^{\infty} \frac{1}{k!+b_k}$ is a rational number.

Before the proof, for every integer $n\geq0$ we denote
\[ \alpha_n := \sum_{k=n+1}^{\infty} \frac{1}{2^k+5} \]
and
\[ \beta_n := \sum_{k=n+1}^{\infty} \frac{1}{2^k+1}. \]

\begin{lemma}\label{lemma1}
For every integer $n\geq0$ we have $\alpha_n<\beta_n$.
\end{lemma}

\begin{proof}[Proof of Lemma \ref{lemma1}]
We have
\begin{align*}
\alpha_n & = \frac{1}{2^{n+1}+5} + \sum_{k=n+2}^{\infty} \frac{1}{2^k+5} \\
& < \frac{1}{2^{n+1}+1} + \sum_{k=n+2}^{\infty} \frac{1}{2^k+5} \\
& \leq \frac{1}{2^{n+1}+1} + \sum_{k=n+2}^{\infty} \frac{1}{2^k+1} = \beta_n,
\end{align*}
which completes the proof.
\end{proof}

\begin{lemma}\label{lemma2}
For every integer $n\geq0$ we have $\beta_{n+1}\geq \alpha_{n+1} + \frac{1}{(2^{n+1}+1)(2^{n+1}+2)}$.
\end{lemma}

\begin{proof}[Proof of Lemma \ref{lemma2}]
By Lemma \ref{lemma1} we have
\begin{align*}
\beta_{n+1} - \alpha_{n+1}
& = \frac{1}{2^{n+2}+1} - \frac{1}{2^{n+2}+5} + \beta_{n+2} - \alpha_{n+2} \geq \frac{1}{2^{n+2}+1} - \frac{1}{2^{n+2}+5} \\
& = \frac{4}{(2^{n+2}+1)(2^{n+2}+5)} = \frac{1}{4^{n+1} + 3\cdot 2^{n+1} + 5/4} \\
& > \frac{1}{4^{n+1} + 3\cdot 2^{n+1} + 2} = \frac{1}{(2^{n+1}+1)(2^{n+1}+2)},
\end{align*}
which completes the proof.
\end{proof}


\begin{proof}[Proof of Theorem \ref{maintheorem}]
By Lemma \ref{lemma1} we have $\alpha_0<\beta_0$, so there exists a rational number $x$ such that $\alpha_0\leq x\leq \beta_0$.
We recursively define the sequence $(b_k)_{k=1}^{\infty}$ and, simultaneously, inductively verify the property
\begin{equation}\label{property}
\sum_{k=1}^{n} \frac{1}{2^k+b_k} + \alpha_n \leq x \leq \sum_{k=1}^{n} \frac{1}{2^k+b_k} + \beta_n
\end{equation}
for every integer $n\geq 0$.
Property \eqref{property} is trivially satisfied for $n=0$ by the assumption $\alpha_0\leq x\leq\beta_0$ and since a sum over an empty set is interpreted as $0$.
Take an integer $n\geq0$ and, if $n\geq1$, suppose that the terms $b_m$ for all integers $m$ satisfying $1\leq m\leq n$ have already been defined.
By \eqref{property} we have
\begin{equation}\label{property2}
\sum_{k=1}^{n} \frac{1}{2^k+b_k} + \frac{1}{2^{n+1}+5} + \alpha_{n+1} \leq x \leq \sum_{k=1}^{n} \frac{1}{2^k+b_k} + \frac{1}{2^{n+1}+1} + \beta_{n+1}.
\end{equation}
Recursively define $b_{n+1}$ to be the largest $c\in\{1,2,3,4,5\}$ such that
\[ x \leq \sum_{k=1}^{n} \frac{1}{2^k+b_k} + \frac{1}{2^{n+1}+c} + \beta_{n+1}. \]
By this definition we have
\begin{equation}\label{ineq1}
x \leq \sum_{k=1}^{n} \frac{1}{2^k+b_k} + \frac{1}{2^{n+1}+b_{n+1}} + \beta_{n+1}.
\end{equation}
If $b_{n+1}=5$, then \eqref{property2} and \eqref{ineq1} together give
\[ \sum_{k=1}^{n+1} \frac{1}{2^k+b_k} + \alpha_{n+1} \leq x \leq \sum_{k=1}^{n+1} \frac{1}{2^k+b_k} + \beta_{n+1} \]
and the recursive step is complete.
Otherwise, if $b_{n+1}\in\{1,2,3,4\}$, then the maximality from its definition gives
\[ x > \sum_{k=1}^{n} \frac{1}{2^k+b_k} + \frac{1}{2^{n+1}+b_{n+1}+1} + \beta_{n+1}, \]
which, in combination with Lemma \ref{lemma2} followed by $b_{n+1}\geq1$, implies
\begin{align*}
x & > \sum_{k=1}^{n} \frac{1}{2^k+b_k} + \frac{1}{2^{n+1}+b_{n+1}+1} + \alpha_{n+1} + \frac{1}{(2^{n+1}+1)(2^{n+1}+2)} \\
& = \sum_{k=1}^{n} \frac{1}{2^k+b_k} + \frac{1}{2^{n+1}+b_{n+1}} + \alpha_{n+1} - \frac{1}{(2^{n+1}+b_{n+1})(2^{n+1}+b_{n+1}+1)} + \frac{1}{(2^{n+1}+1)(2^{n+1}+2)} \\
& \geq \sum_{k=1}^{n} \frac{1}{2^k+b_k} + \frac{1}{2^{n+1}+b_{n+1}} + \alpha_{n+1},
\end{align*}
so that
\[ \sum_{k=1}^{n+1} \frac{1}{2^k+b_k} + \alpha_{n+1} \leq x. \]
From the last equation and \eqref{ineq1} we conclude
\[ \sum_{k=1}^{n+1} \frac{1}{2^k+b_k} + \alpha_{n+1} \leq x \leq \sum_{k=1}^{n+1} \frac{1}{2^k+b_k} + \beta_{n+1} \]
and the recursive step is complete in this case as well.

Since the series $\sum_{k=1}^{\infty} \frac{1}{2^k+5}$ and $\sum_{k=1}^{\infty} \frac{1}{2^k+1}$ converge, we have that their tails converge to $0$, so $\lim_{n\to\infty}\alpha_n=0$ and $\lim_{n\to\infty}\beta_n=0$.
Using this, taking the limits of all three expressions in \eqref{property}, and applying the squeeze theorem, we conclude
\[ \sum_{k=1}^{\infty} \frac{1}{2^k+b_k} = x, \]
which is a rational number by our choice of $x$.
\end{proof}

\end{document}