I struggled with the following question, and would like to know how to do it correctly.
The exponential function is defined by $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$. Assuming that $\lim_{N\to\infty} \sum_{n=0}^N \frac{x^n}{n!}$ converges uniformly on every interval of the form $[-M,M]$, show that $\int_0^x e^t\ dt = e^x - 1$.
My strategy that I believe was incorrect was to say since $f_n \rightarrow f$ is uniformly convergent then $\int f_n=\int f$
Then I said because of this, $\lim_{n\to\infty}\sum_{n=0}^{\infty} \int_{0}^{x} \frac{t^n}{n!}dt=\int_{0}^{x}e^{x}dx=e^{x}-1$ but I realize I have not really "shown" anything.
Well, what is the correct way to solve this problem?
$\int_0^x\sum_{n=0}^N \frac{t^n}{n!}dt=\sum_{n=0}^N \frac{x^{n+1}}{(n+1)!}=\sum_{n=0}^{N+1}\frac{x^n}{n!}-1\to e^x-1$
The convergence is uniform on any finite interval.