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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!

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

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

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