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Using the definition of uniform convergence prove that the exponential series $\sum_{k=0} ^\infty \frac{x^k}{k!}$ converges uniformly on any finite subinterval of $\mathbb{R}$.

The sequence of functions $(f_n)_{n\geq 1}$ on $A \subset \mathbb{R}$ converges uniformly to $f:A \to \mathbb{R}$ if $\forall \varepsilon > 0 $ there exist $N$ such that $\forall x \in A, \forall n > N$, $|f_n(x)-f(x)| < \varepsilon$.

So far I have let $f_n(x)=\sum_{k=0} ^n \frac{x^k}{k!}$, let $f(x)=e^x$. For $(a,b) \subset \mathbb{R}$, $|f_n(x)-f(x)| = |\sum_{k=0} ^n \frac{x^k}{k!} - e^x|$, but then I am not sure how to go from here. I wanted to turn $|f_n(x)-f(x)|$ into something that approaches $0$ as $n \to \infty$, so that it can be less than $\varepsilon$.

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  • $\begingroup$ $|f_n(x)-f(x)| = |\sum_{k=n+1} ^\infty \frac{x^k}{k!}|$ Can figure out an upper limit for that sum when $|x| < M$ for some constant $M$? $\endgroup$ – Paul Sinclair Nov 8 '15 at 22:54
  • $\begingroup$ @PaulSinclair If I let $M=max\{|a|,|b|\}$, so $|x|<M$. Then $|f_n(x)-f(x)| \leq |\sum_{k=n+1} ^\infty \frac{M^k}{k!}|$, but I still can't see how to find the upper limit. $\endgroup$ – Lucy Nov 8 '15 at 23:04
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    $\begingroup$ That is the upper limit. It holds for all $x$ which is what you need for uniform continuity. What happens to $\sum_{k=n+1} ^\infty \frac{M^k}{k!}$ as $n \to \infty$? $\endgroup$ – Paul Sinclair Nov 8 '15 at 23:07
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    $\begingroup$ @PaulSinclair Can I say that as $n \to \infty$ , $k \to \infty$ so the sum tends to $0$? But would that mean $\frac{M^k}{k!} \to \infty$ as well? $\endgroup$ – Lucy Nov 8 '15 at 23:11
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    $\begingroup$ Yes. If it didn't, the full sum would not converge to $e^M$ like we know it does.(you changed your comment since I posted this - I mean that the sum $\to 0$ as $n \to \infty$, as for the other $M^k/k! \to 0$ as $k \to \infty$ $k!$ outstrips $M^k$. $\endgroup$ – Paul Sinclair Nov 8 '15 at 23:16
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You can prove a much more general result: Consider the series $\sum_{k=0}^\infty a_kx^k$ of radius of convergence $\mathcal R$. Then it converge uniformly on any compact $K\subset ]-\mathcal R,\mathcal R[$. Indeed, let $0<r<\mathcal R$ and let denote $S_n(x)=\sum_{k=0}^n a_kx^k$. In particular if $m>n$ and $x\in [-r,r]$ $$|S_m(x)-S_n(x)|\leq \sum_{k={n+1}}^m |a_k||x|^k\leq\sum_{k=n+1}^m|a_k|r^k\implies \sup_{x\in [-r,r]}|S_n(x)-S_m(x)|<\sum_{k=n+1}^m |a_k|r^k$$ and thus, $(S_n)$ is a sequence of continuous function of Cauchy for the norm of the maximum. Therefore, it converge uniformly.

Since the radius of convergence of your series is $\mathbb R$, you got your result.

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