TITLE: A dominated convergence theorem applied to $e$ number definition
QUESTION [9 upvotes]: I want to show that:
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^\infty \frac{1}{k!}.$$
By the binomial theorem
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \lim_{n\to\infty}\sum_{k=0}^n \frac{1}{k!} \frac{n(n-1)\cdots(n-k+1)}{n^k}=\lim_{n\to\infty}\sum_{k=0}^n f_n(k).$$
Here I have problems with limits, since the limit of the sum depends on $n$ and also the summation terms. I tried to fix it with the following:
Let $g_n:=f_n\chi_{P_n}\colon \mathbb{N}\to \mathbb{R}$ measurable functions in the counting measure $\mu$ in $\mathbb{N}$ 
we have that $$g_n(k)=f_n(k)\chi_{P_n}(k)\to \frac{1}{k!}$$
for each $k\in\mathbb{N}$ and $P_n=\{1,\ldots,n\}$. But
$$|g_n(k)|\leq \frac{1}{k!}:=f(k)\quad\text{and}\quad \int_{\mathbb{N}}f(k)d\mu(k)=\sum_{k=0}^\infty \frac{1}{k!}<3<\infty.$$
So by the Lebesgue Dominated Convergence Theorem,
\begin{align}
\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n&=\lim_{n\to\infty}\sum_{k=0}^n f_n(k)\\
&=\lim_{n\to\infty}\int_{P_n}f_n(k)d\mu(k)\\
&=\lim_{n\to\infty}\int_{\mathbb{N}}\chi_{P_n}(k) f_n(k)d\mu(k)\\
&=\lim_{n\to\infty}\int_{\mathbb{N}}g_n(k)d\mu(k)\\
&=\int_{\mathbb{N}}\lim_{n\to\infty}g_n(k)d\mu(k)\\
&=\int_{\mathbb{N}}f(k)d\mu(k)\\
&=\int_{\mathbb{N}}\frac{1}{k!}d\mu(k)\\
&=\sum_{k=0}^\infty \frac{1}{k!}.
\end{align}
Am I right?

REPLY [2 votes]: Your second and third equalities in the bottom chain are incorrect, they should be 
$$
= \lim \int_{P_n} f_n d\mu = \lim \int_{\mathbb{N}} \chi_{P_n} f_n d\mu
$$
But the rest of the proof goes through if you correct this small typo. Proof looks good.