Problem regarding convergency: $\sum_{k=0}^{\infty}\frac{x^k}{k!} \to e^x$ 
Question. Let $X$ be the space of all polynomials in one variable,with real coefficients. If $p=a_0+a_1x+\dots+a_nx^n\in X$, define $$|p|=|a_0|+|a_1|+\dots+|a_n|,$$ which gives the metric $d(p,q)=|p-q|$ on $X$. 
Does the metric space $X$ is complete?

Now, this question has been answered 'false' here.
According to that answer let $p_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}$
...
I know this is due to the fact of Taylor series (theorem) $$\sum_{k=0}^{\infty} \frac{x^k}{k!}= e^x  ,\  \text{*as*} \ n \longrightarrow \infty \  ,\forall \ x \in \mathbb{R}: ~p_n \to e^x$$ 
But still I have some questions:

  
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*When we say $p_n \to e^x$ as $n \to \infty$.then In which space this convergency is happening...?? I mean w.r.to which norm? Does the space here is $X=C(\mathbb{R}):= \text{Space of continuous functions on}~ \mathbb{R}$ ...But then which norm?? I mean "sup-norm" shouldn't work here as $e^x$ is not bounded...!!
  
*Now how does this convergency (i.e. $p_n\to e^x$) remains independent of norm put on the space?

Please help me to clarify these confusions. Correct me if I'm wrong. Thank you.

The statement that $p_n$ converges to $e^{x}$ and hence it does not converge in the given  space is just suppose to be a hint and not a complete proof. No norm is used in saying that $p_n$ converges to $e^{x}$. You have to give a rigorous proof along the following lines: if a sequence of polynomials converges  in the given norm then the coefficients converge. So if $p_n$ converges to  a polynomial $p$ then  its  coefficients have to be $\frac 1 {k!}$, $k=0,1,2,...$. In particular no coefficient can be $0$ which contradicts the fact that $p$ is a polynomial. To be more explicit suppose $p$ has degree $m$. Since each each $p_n$ with $n >m$ has $\frac 1 {(m+1)!}$ as the $m+1$-st coefficient the same must be true of $p$ but $p$ does not have this term.
