# Is the Taylor series uniformly convergent on $\Bbb R$?

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 $R$?

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

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

• 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 Jan 11, 2017 at 23:05
• @zhw: I really appreciate if you explain more.
– SAm
Jan 12, 2017 at 7:17

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...

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