Using the induction method:

$(\forall P)[[P(0) \land ( \forall k \in \mathbb{N}) (P(k) \Rightarrow P(k+1))] \Rightarrow ( \forall n \in \mathbb{N} ) [ P(n) ]]$

Why this proof is wrong?

$P(x)\equiv (\displaystyle\lim\limits_{a \to x}\sum_{i=0}^{a}\frac{1}{i!}\in\mathbb{Q})$

**Basis**

$P(0)\equiv$ **True** ?

$P(0)\equiv(1\in\mathbb{Q})\equiv True$

**Induction**

$k \in \mathbb{N}$

$P(k)\equiv$ **True** $\implies P(k+1)\equiv$ **True**?

$P(k+1)\equiv$
$(\displaystyle\lim\limits_{a \to k}\sum_{i=0}^{a+1}\frac{1}{i!}\in\mathbb{Q}) \implies (\displaystyle\lim\limits_{a \to k}\sum_{i=0}^{a}\frac{1}{i!}+\lim\limits_{a \to k}\frac{1}{(a+1)!}\in\mathbb{Q})\implies (P(k)+\frac{1}{(k+1)!} \in \mathbb{Q})\implies$ **True**

So, $P(x)$ is **true** for all $x \in \mathbb{N}$.

But,

$\lim\limits_{x \to \infty}{P(x)}\equiv (\displaystyle\lim\limits_{x \to \infty}\lim\limits_{a \to x}\sum_{i=0}^{a}\frac{1}{i!}\in\mathbb{Q})\equiv \lim\limits_{x \to \infty}\sum_{i=0}^{a}\frac{1}{i!}\in\mathbb{Q}\implies e \in \mathbb{Q}$. But we know $e$ don't belongs $\mathbb{Q}$.

So I would like to know, why this fake proof doesn't work, why run through all integers in $x$ until the last one, infinity, the formula fails.

**Edit:**
I had include limits to extend the discussion.