Suppose you have:

$S$ a set of increasing sequences with elements in $\mathbb{N} \setminus \{0,1\}$. That is $\forall (u_n)_{n \in \mathbb{N}} \in S$, $u_{n+1}-u_n \geq 0$. Now suppose you have $(a_n)_{n \in \mathbb{N}} \in S$ and define $$x_n = \frac{1}{a_0} + \frac{1}{a_0 a_1} + \cdot \cdot \cdot \frac{1}{a_0 a_1 \cdot \cdot \cdot a_n}$$

Show that the function who for each $a_n \in S$ associates $\lim x_n$ is bijective.

I can see that by construction is it surjective, but no idea how to proceed with injectiveness.


Suppose that $(a_n)$ and $(b_n)$ have the same image, but $(a_n)\neq (b_n)$. Consider the smallest integer $N$ such that $a_{N+1}\neq b_{N+1}$. We assume without loss of generality that $a_{N+1}>b_{N+1}$.

We have $\sum_{k=0}^\infty\frac{1}{\prod_{j=0}^k a_j}=\sum_{k=0}^\infty\frac{1}{\prod_{j=0}^k b_j}$ and since $a_0=b_0, \ldots, a_{N}=b_{N}$, the equality simplifies as $$\sum_{k=N+1}^\infty\frac{1}{\prod_{j=N+1}^k a_j}=\sum_{k=N+1}^\infty\frac{1}{\prod_{j=N+1}^k b_j}$$

Since $a_{N+1}$ and $b_{N+1}$ are integers, $a_{N+1}>b_{N+1}$ implies $a_{N+1}\geq b_{N+1}+1$. Monotony of $(a_n)$ and the previous inequality yield $$\sum_{k=N+1}^\infty\frac{1}{\prod_{j=N+1}^k a_j}\leq \sum_{k=N+1}^\infty\frac{1}{(b_{N+1}+1)^{k-N}}=\frac{1}{b_{N+1}} $$

Hence $$\sum_{k=N+1}^\infty\frac{1}{\prod_{j=N+1}^k b_j}\leq \frac{1}{b_{N+1}}$$ which is absurd.


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