# Prove the series does not converge uniformly

Prove the series $\sum\limits_{n=0}^\infty\frac{x^n}{n!}$ does not converge uniformly on $\mathbb{R}$.

So what I am thinking is that the pointwise summation is $e^x$ and that I need to show there is no $n\geq N$ so that $|\sum\limits_{n=0}^\infty\frac{x^n-e^xn!}{n!}| \leq \epsilon$, for all x . We haven't covered integration; if someone could give me a hint on how to show the series does not converge uniformly, that we be great. Thanks in advance!

• The terms of the sum do not converge uniformly to $0$ on $\Bbb R$. – David Mitra Jan 23 '14 at 14:14

## 1 Answer

The difference between the function and the partial sum $$e^x-\sum\limits_{n=0}^k\frac{x^n}{n!}$$ is bounded? Answer: no. Why?

• Because if we fixed $k$ and let $x -> \infty$ then $e^x$ grows much faster than the partial sum? Then the difference would continue to grow as x approaches infinity. – Fluke_of_Luke Jan 23 '14 at 15:12
• True. And the difference between the function and the partial sum cannot be small. BTW, edit required in your post, because $\sum\limits_{n=0}^\infty\frac{x^n-e^xn!}{n!}=-\infty$. – Martín-Blas Pérez Pinilla Jan 23 '14 at 15:48
• @Fluke_of_Luke ehm why e^x grows much faster than the partial sum? It seems to me that it should be justified.. – Surfer on the fall Jun 14 '16 at 9:59