Given that $\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e$, show that $1+\frac{1}{1!}+\frac{1}{2!}+\cdots=e$. Problem
Given that $\lim\limits_{n \to \infty}\left(1+\dfrac{1}{n} \right)^n = e$, show that $1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots=e$.
Proof
By binomial theorem, we have 
\begin{align*}\left(1+\dfrac{1}{n}\right)^n&=\sum_{k=0}^{k=n}\binom{n}{k}1^k \left(\frac{1}{n}\right)^{n-k}\\
&=\sum_{k=0}^{k=n}\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right).
\end{align*}
Thus, on one hand, $$\left(1+\dfrac{1}{n}\right)^n \leq \sum_{k=0}^{k=n}\frac{1}{k!}.\tag1$$
Take the limits as $n \to \infty$ on the both sides of $(1)$. We have $$e=\lim_{n \to \infty}\left(1+\dfrac{1}{n}\right)^n\leq \varliminf_{n \to \infty} \sum_{k=0}^{k=n}\frac{1}{k!}.\tag2$$
On the other hand, take a positive integer $m$ such that $m<n$ and fix it. We have 
$$\left(1+\dfrac{1}{n}\right)^n \geq \sum_{k=0}^{k=m}\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right).\tag3$$
Likewise,take the limits as $n \to \infty$ on the both sides of $(3)$. We have 
$$e=\lim_{n \to \infty}\left(1+\dfrac{1}{n}\right)^n\geq \sum_{k=0}^{k=m}\frac{1}{k!}.\tag4$$
Take the limits as $m \to \infty$ on the both sides of $(4)$. We have 
$$e \geq \varlimsup_{m \to \infty}\sum_{k=0}^{k=m}\frac{1}{k!}.\tag 5$$
Combine $(2)$ and $(5)$. It follows that $$\lim_{n \to \infty}\sum_{k=0}^{k=n}\frac{1}{k!}=e.$$
Please correct me If I'm faulty. Hope to see other solutions.

Another more elegant proof
At the beginning, let's recall an unusual fact, named Tannery's limit theorem, which states that

Let $S(n)=\sum\limits_{k=0}^{n}a_k(n)$.
If the following conditions are all satisfied:

*

*For any $k$, $\lim\limits_{n \to \infty}a_k(n)=a_k$;

*for any $k$，there exists a $M_k>0$ independent of $n$ such that     $|a_k(n)|\leq M_k$;

*$\sum\limits_{k=0}^{\infty}M_k<\infty$,

then $$\lim\limits_{n \to\infty}S(n)=\sum\limits_{k=1}^{\infty}a_k.$$

As for the present problem, we may denote $$S(n)=\left(1+\frac{1}{n}\right)^n,$$ and$$a_k(n)=\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right).$$ Thus, we may verify that:

*

*For any $k$, $\lim\limits_{n \to \infty}a_k(n)=\dfrac{1}{k!}=a_k$;

*for any $k$，$|a_k(n)|\leq \dfrac{1}{k!}=M_k$;

*$\sum\limits_{k=0}^{\infty}\dfrac{1}{k!}=\sum\limits_{k=0}^{\infty}M_k<2+\sum\limits_{k=2}^{\infty}\dfrac{1}{2^{k-1}}=3<\infty$.

It follows that $$e=\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=\lim\limits_{n \to \infty}S(n)=\sum\limits_{k=0}^{\infty}\dfrac{1}{k!}=\sum\limits_{k=0}^{\infty}a_k.$$
