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:

  
*
  
*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.

Proving that a metric space is not complete means showing that there exists a Cauchy sequence (https://en.wikipedia.org/wiki/Cauchy_sequence) of its elements which does not have a limit in this space. 
So basically, you first want to prove that the elements of your sequence become arbitrary close (that for $ \forall \epsilon > 0, \exists N: \forall n, m > N, |p_n - p_m| < \epsilon  $).
Also, since your metric is defined only on  polynomials, you can not directly prove that $ p_n $ converges to $ e^x $ (because ($e^x - p_n$) is not a polynomial). The easiest way around is to suppose that the given sequence has some polynomial as its limit, and come to a contradiction. (as it is done in the above comment) 
