# Proving unif. conv. of $\lim_{m \rightarrow \infty} \sum_{k=0}^m \frac{1}{k!} \left( 1 - \frac{1}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right)$

I am currently working on trying to prove that $$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^\infty \frac{1}{k!}.$$

Here is my progress so far.

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n = \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{n!}{k! (n-k)!} \left(\frac{1}{n} \right)^n$$

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n = \lim_{n \rightarrow \infty} \sum_{k=0}^n\frac{1}{k!} \cdot \frac{n}{n} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac{n-(k-1)}{n}$$

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n = \lim_{n \rightarrow \infty} \sum_{k=0}^n\frac{1}{k!} \left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right )$$

Since for all $$k > n$$, we have $$\left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right ) = 0$$, since one of the factors of the form $$\left( 1 - \frac{i}{n} \right )$$ is guaranteed to be zero when $$k > n$$, we can make the sum go to infinity without affecting the result.

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n = \lim_{n \rightarrow \infty} \sum_{k=0}^\infty \frac{1}{k!} \left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right )$$

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n = \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \sum_{k=0}^m \frac{1}{k!} \left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right )$$

At this point, I'd like to be able to swap the limits, which would allow the summation to reduce to $$\sum \frac{1}{k!}$$ as desired. To my knowledge, this is only possible when the conditions of the Moore-Osgood Theorem are satisfied, which say that if $$f(m,n) = \sum_{k=0}^m \frac{1}{k!} \left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right)$$ then we must have either $$\lim_{m\rightarrow \infty} f(m,n)$$ or $$\lim_{n\rightarrow \infty} f(m,n)$$ be uniformly convergent in order to be able to swap the $$m \rightarrow \infty$$ and $$n \rightarrow \infty$$ limits above.

If we define $$g(n) = \sum_{k=0}^n \frac{1}{k!} \left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right )$$, it is fairly evident that $$\lim_{m \rightarrow \infty} f(m,n) = g(n)$$ pointwise, but only uniform convergence would be sufficient for the Moore-Osgood Theorem.

To show uniform convergence here, one would need to prove that $$\forall \varepsilon > 0 : \exists M : \forall m \geq M : \forall n$$ we have that $$|f(m,n) - g(n)| < \varepsilon$$. I managed to deduce that

$$|f(m,n) - g(n)|= \sum_{k=m+1}^n \frac{1}{k!} \left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right ) \text{if } m< n \text{ and } 0 \text{ otherwise.}$$

At this point I feel like I must be really close, but my brain is a bit fried trying to figure out how to show that this is less than $$\varepsilon$$ for any arbitrary choice of $$\varepsilon > 0$$ under all possible values of $$m \geq M$$ and $$n$$ for a chosen $$M$$. Is this even possible and could anyone provide a possible next step or hint of where to go from here?

To avoid confusion, the expression $$\left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right )$$ should be replaced with

$$\alpha_{kn} = \begin{cases}1, &k=0,1\\ \left( 1 - \frac{1}{n} \right )\left( 1 - \frac{2}{n} \right ) \cdots \left( 1 - \frac{k-1}{n} \right ), &2\leqslant k \leqslant n\\0,&k > n\end{cases},$$ and using the binomial theorem we have

$$x_n :=\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^k}\\= 1+1 + \frac{1}{2!} \left(1 - \frac{1}{n} \right)+ \ldots +\frac{1}{n!} \left(1 - \frac{1}{n} \right)\cdots \left(1 - \frac{n-1}{n} \right)=\sum_{k=0}^n \frac{\alpha_{kn}}{k!}$$

You are trying to prove that $$\lim_{n \to \infty} x_n = \sum_{k=0}^\infty \frac{1}{k!} = e$$. Since you prefer viewing this in terms of a double limit, we define (using your notation)

$$f(m,n) = \sum_{k=0}^m \frac{\alpha_{kn}}{k!},$$

and since $$\lim_{n\to \infty}\alpha_{kn} = 1$$ for $$k\leqslant m$$ it follows that

$$\tag{1}\lim_{n \to \infty}f(m,n) = \sum_{k=0}^m \frac{1}{k!}$$

Also, since $$0 \leqslant \alpha_{kn}\leqslant 1$$, it follows for all $$m < n$$ that

$$\tag{2}f(m,n) \leqslant x_n = f(n,n) \leqslant \sum_{k=1}^n \frac{1}{k!}$$

Hence, one iterated limit of $$f(m,n)$$ is
$$\tag{3}\lim_{m \to \infty}\lim_{n \to \infty}f(m,n) = \lim_{m \to \infty}\sum_{k=0}^m \frac{1}{k!} = \sum_{k=0}^\infty \frac{1}{k!}$$

No swapping of limits is needed to prove the desired result since from (2) we have

$$\lim_{n \to \infty} f(m,n) \leqslant \liminf_{n \to \infty}x_n \leqslant \limsup_{n \to \infty}x_n \leqslant \lim_{n \to \infty} \sum_{k=0}^n\frac{1}{k!} = \sum_{k=0}^\infty \frac{1}{k!},$$

and using (3) it follows that

$$\sum_{k=0}^\infty \frac{1}{k!} \leqslant \liminf_{n \to \infty}x_n \leqslant \limsup_{n \to \infty}x_n \leqslant \sum_{k=0}^\infty \frac{1}{k!}$$

This proves that the limit of $$x_n$$ exists and

$$\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = \lim_{n \to \infty}x_n = \sum_{k=0}^\infty \frac{1}{k!}$$

Regarding uniform convergence

Even though it is not needed, we can show that $$f(m,n)$$ converges uniformly as $$m \to \infty$$.

Since $$\alpha_{kn} = 0$$ for $$k > n$$, we have

$$\lim_{m \to \infty}f(m,n) = \sum_{k=0}^\infty \frac{\alpha_{kn}}{k!} = \sum_{k=0}^n \frac{\alpha_{kn}}{k!}$$

Hence, for $$n \leqslant m$$

$$\left|\sum_{k=m+1}^\infty\frac{\alpha_{kn}}{k!} \right|= 0$$

and for all $$n > m$$

$$\left|\sum_{k=m+1}^\infty\frac{\alpha_{kn}}{k!} \right| < \sum_{k=m+1}^\infty \frac{1}{k!}$$

Since the RHS is the tail of a convergent series that does not depend on $$n$$ uniform convergence follows.