A sequence defined by another sequence Let $(a_n)_{n\ge 1}$ be a sequence of positive integers greater or equal than $2$.
Consider the sequence $b_n=1-\frac{1}{a_1}+\frac{1}{a_1 a_2}-...+(-1)^n\frac{1}{a_1 a_2...a_n}$, $n=1,2,3...$.
Prove that if $(a_n)_{n\ge 1}$ is unbounded, then $\lim\limits_{n\to \infty}b_n \in \mathbb{R}-\mathbb{Q}$.
This is the second task of this problem. The first one was to prove that $(b_n)_{n\ge 1}$ converges. I could solve that one by proving that $(b_{2n})_{n\ge 1}$ and $(b_{2n+1})_{n\ge 1}$ both converge and have the same limit, but I have no clue how to prove that the limit is irrational if $(a_n)_{n\ge 1}$ is unbounded.
 A: You have already shown that $(b_{2n})$ and $(b_{2n+1})$ converge
to the same limit ; denote this limit by $l$.
Since
$$
\begin{array}{c}
b_{2n}-b_{2n+2}=\frac{a_{2n+2}-1}{a_1a_2\ldots a_{2n+2}} \gt 0, \\
b_{2n+3}-b_{2n+1}=\frac{a_{2n+3}-1}{a_1a_2\ldots a_{2n+3}} \gt 0
\end{array}
\tag{1}
$$
we see that $(b_{2n})$ is decreasing and $(b_{2n+1})$ and increasing. It follows that
$$
b_{2n+1} \lt l \lt b_{2n} \tag{2}
$$
Now, suppose that $l$ is a rational number, $l=\frac{p}{q}$ where $p$
and $q$ are integers and $q\gt 0$. Let
$$u_n=b_n(a_1a_2a_3\ldots a_n)\tag{3}$$
Then $u_n$ is an integer, and
$$
\frac{u_{2n+1}}{a_1a_2\ldots a_{2n+1}} \lt \frac{p}{q}
\lt \frac{u_{2n}}{a_1a_2\ldots a_{2n}} \tag{4}
$$
Next, in view of (4) let us define
$$
\begin{array}{lcl}
d_1&=&(a_1a_2\ldots a_{2n+1})p-u_{2n+1}q, \\
d_2&=&u_{2n}q-(a_1a_2\ldots a_{2n})p \tag{5}
\end{array}
$$
It follows from (4) that both $d_1$ and $d_2$ are positive integers, so that $d_1\geq 1$ and $d_2\geq 1$. Now, if we set a linear combination that eliminates $p$  :
$$
\begin{array}{lcl}
d_1+a_{2n+1}d_2 &=& (a_{2n+1}u_{2n}-u_{2n+1})q \\
 &=& (a_1a_2\ldots a_{2n+1})(b_{2n}-b_{2n+1})q \\
 &=& q.
\end{array}
$$
So $q=d_1+a_{2n+1}d_2 \geq 1+a_{2n+1}$. A similar argument (using
$b_{2n+1} \lt l \lt b_{2n+2}$) shows that $q\geq 1+a_{2n+2}$. So if $(a_n)$ is unbounded, $q$ is an integer greater than all integers and this is absurd. This finishes the proof.
