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How can we Prove that $\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$ does not converge uniformly on $\mathbb{R}$? By using weierstrass-M test, it is easy to show that this series converges uniformly on a compact interval. Is it true to say that:\ By contradiction, suppose $\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$ converges uniformly to $e^x$, then $(s_{n+1}-s_n)$ converges uniformly to zero, however, $\dfrac{x^{n+1}}{(n+1)!}$ is not bounded on $R$?

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  • $\begingroup$ Because $\frac{n^n}{n!}>1$ for every $n$. $\endgroup$ Jan 11, 2017 at 23:02

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Hint: If $\sum f_n$ converges uniformly on a set $E,$ then $\sup_E|f_n|\to 0.$

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    $\begingroup$ It follows that this is not only the case for the series for $e^x$, but rather for every Taylor series unless it is in fact a polynomial $\endgroup$ Jan 11, 2017 at 23:05
  • $\begingroup$ @zhw: I really appreciate if you explain more. $\endgroup$
    – SAm
    Jan 12, 2017 at 7:17
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Suppose that the series converges uniformly on $\Bbb R$. In particular, taking $\epsilon = 1$, $\exists n_0$ s.t. for $n\ge n_0$, $x\in\Bbb R$: $$\left|\sum_{k=0}^{n}\frac{x^k}{k!} - e^x\right|\le 1.$$ But this is impossible because at least two reasons:

(1) When $x\to+\infty$ the exponential growths faster than any polynomial.

(2) When $x\to-\infty$ the exponential $\to 0$ while the polynomial...

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  • $\begingroup$ Thank you so much. But how can I show that if $\sup\mid (s_n-e^x)\mid <1$, then $\lim_{n\to\infty}\sup\mid (s_n-e^x)\mid=0$? $\endgroup$
    – SAm
    Jan 12, 2017 at 8:51
  • $\begingroup$ @SAm, is false that $\lim_{n\to\infty}\sup∣(s_n−e^x)∣=0$. $\endgroup$ Jan 12, 2017 at 9:46
  • $\begingroup$ I think by contradiction we can suppose this lim is zero $\endgroup$
    – SAm
    Jan 12, 2017 at 19:05
  • $\begingroup$ @SAm, uniform convergence $\implies$ bounded difference, but because of (1), (2) the difference can't be bounded. $\endgroup$ Jan 12, 2017 at 21:00

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