Prove that the series

$\Large{exp(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}}$

does not uniform converge on $\mathbb{R}$, although the limit function
(convergence radius $R=+\infty$) is continous on $\mathbb{R}$.

Question

Prove that the series

$\Large{exp(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}}$

does not uniform converge on $\mathbb{R}$, although the limit function 
(convergence radius $R=+\infty$) is continous on $\mathbb{R}$.

kinggeorge · Answer

From what it seems like to me, we need to show that the sequence of partial sums is not uniformly convergent. How we would show that, I'm not entirely sure yet.

anonymous · Answer

hmm.. maybe Mr Elias has idea for it..

anonymous · Answer

Let 
$$s_n(x) =\sum_1^n \frac {x^i}{i!}
$$
Suppose the the convergence is uniform. That means that the is N such that
for n >= N such that
$$
|s_n(x) -e^x| < 1, \text { for all } x\
\text { in particular }
|s_N(x) -e^x| < 1, \text { for all } x\

$$
This is not possible since 
$$
\lim_{x \to \infty} S_N(x)= + \infty
$$

anonymous · Answer

I still have to make a better proof.

anonymous · Answer

Here how to make it works.
Let N odd, then
$$
\lim_{n\to - \infty} S_N(x)= -\infty \
| -\infty -0 | <1
$$
Which is a contradiction

anonymous · Answer

Mr Elias it looks good but if you want to make better proof i will be happier, because my mentor is very mean about giving points.. maybe George can say something additional

anonymous · Answer

It is fixed now, See my previous post.

anonymous · Answer

ok thank you very much Mr Elias,

anonymous · Answer

yw