I am trying to solve the following question . Let $r\in (0,1)$ and $m\in \mathrm{N}$. Then there exists a sequence $(a\_n)\_{n\in \mathrm{N}}$ in $\mathrm{N}$ such that $$ \prod\_{n=1}^{\infty} \left ( 1- \dfrac{m}{a\_n +m} \right ) =r.$$
Is it possible that $ (a\_n)\_{n\in \mathrm{N}}$ be chosen strictly increasing?



Sequence $(a_n)$ for which $ \prod_{n=1}^{\infty} \left ( 1- \dfrac{m}{a_n +m} \right ) =r.$
By simple transformations your claim is equivalent to 
the following: 


For each $M\in \mathbb{N}$ and $s>0$ there is a strictly increasing sequence $(a\_n)\in\mathbb{N}^\infty$, such that
$$ \sum \limits\_{n>0} \log \left( 1+ \frac{m}{a\_n}\right) ~=~ s.$$
This can be proved by inductive construction of such a sequence.
Define $s\_0=0$, and for $N>0$ choose 
$$ a\_N~:=~ \min \left\{ k\in \mathbb{N}: \log \left( 1+ \frac{m}{k}\right) < s-s\_{N-1} \right\}~~~~(1)$$
Apparently, $a\_N$ is well-defined because the set on the right hand side is not empty.
It follows that 
$$ s\_N~:=~ \sum \limits\_{n=1}^N \log \left( 1+ \frac{m}{a\_n}\right) ~<~ s.$$
Because the sequence $(s\_n)\_{n\in \mathbb{N}} $ is monotonously increasing and dominated by $s$, it converges to a limit $s^\*\leq s$.
It remains to show that $s^\*=s$.


Assume $\epsilon=s-s^\*>0$.
Choose $N$ such that for all $M>N$
$$\log \left( 1+ \frac{m}{a\_M}\right)~<~ \epsilon$$
Apparently, the sequence $(a\_n)\_{n>N}$ is not a sequence of consecutive integers, otherwise $(s\_n)$ would diverge, like the harmonic series.
So there are two consecutive members $a\_M$ and $a\_{M+1}$ with $a\_{M+1}-a\_M\geq 2$. Now one can consider the sequence 
where $a\_{M+1}$ is replaced by $a\_{M+1}-1$. 
Apparently this sequence is also a strictly increasing sequence with a larger limit value than $s^\*$, but still smaller than $s$.
But this contradicts minimality of $a\_{M+1}$ in condition (1) of the construction.


