Proving the convergence of an infinite product

Question

I’m trying to prove the Taylor series of $e$ using binomial expansion:

$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$$

The steps I’ve tried so far are: $$\left(1+\frac{1}{n}\right)^n \\ = 1 + \binom{n}{1}\frac{1}{n} + \binom{n}{2}\frac{1}{n^2} + \binom{n}{3}\frac{1}{n^3} + \dots + \binom{n}{n}\frac{1}{n^n} \\ = 1+\dfrac {n}{1!}\left( \dfrac {1}{n}\right) +\dfrac {n\left( n-1\right) }{2!}\left( \dfrac {1}{n^{2}}\right) + \dfrac {n\left( n-1\right) \left( n-2\right) }{3!}\left( \dfrac {1}{n^{3}}\right) +\dots +\dfrac {n\left( n-1\right) \left( n-2\right) \dots 1}{n!}\left( \dfrac {1}{n^{n}}\right) \\ = 1+\dfrac {1}{1!}+\dfrac {1}{2!}\left( 1-\dfrac {1}{n}\right) +\dfrac {1}{3!}\left( 1-\dfrac {1}{n}\right) \left( 1-\dfrac {2}{n}\right) +\ldots +\dfrac {1}{n!}\left( 1-\dfrac {1}{n}\right) \left( 1-\dfrac {2}{n}\right) \dots \left( 1-\dfrac {n-1}{n}\right) $$ As $n \to \infty$, the last term becomes $$\lim_{n\to\infty} \left(\frac{1}{n!} \prod_{k=1}^n \left(1-\frac{k-1}{n}\right)\right)$$ and the infinite product on the right of the term should be proven to be convergent and equals to $1$. So how do I prove that?

Answer 1

First, all of your equalities seem correct, although I'm not entirely sure why the last term should equal 1 in the limit?

In any case, I think your approach so far works well, I think you just need to keep in mind what the goal is. You start with:

$$ e = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n $$

Which you have shown is equivalent to (and I'm formatting your last step as a sum):

$$ e = \lim_{n \rightarrow \infty} \sum\limits_{i=0}^n \left[\frac{1}{i!}\prod\limits_{j=1}^{i-1}\left(1-\frac{j}{n}\right) \right] $$

I think the key insight here (though bear with me, it's not the most rigorous, but it'll do the trick) is considering what exactly

$$ \prod\limits_{j=1}^{i-1} \left(1 - \frac{j}{n} \right) $$

is. If you go ahead and multiply everything out, (especially considering this is finite), you get that $\exists p_i \in \mathbb{Z}[X]$ (where $\mathbb{Z}[X]$ is the set of polynomials with integer coefficients) such that:

$$ \prod\limits_{j=1}^{i-1} \left(1 - \frac{j}{n} \right) = 1 + \frac{p_i\left(\frac{1}{n}\right)}{n} $$

This must be true, since the constant term of the product must be $1$, and the remaining terms all contain $\frac{1}{n}$, so it can be factored out. The rest is a polynomial ($p_i$, different for each $i$) with integer coefficients evaluated at $\frac{1}{n}$. Now, replacing this into the limit,

\begin{align*} e &= \lim_{n \rightarrow \infty} \sum\limits_{i=0}^n \left[\frac{1}{i!}\prod\limits_{j=1}^{i-1}\left(1-\frac{j}{n}\right) \right] \\ &= \lim_{n \rightarrow \infty} \sum\limits_{i=0}^n \left[\frac{1}{i!}\left(1+\frac{p_i\left(\frac{1}{n}\right)}{n}\right) \right] \\ &= \lim_{n \rightarrow \infty} \sum\limits_{i=0}^n \frac{1}{i!} + \lim_{n \rightarrow \infty} \frac{1}{n}\sum\limits_{i=0}^n \left[\frac{1}{i!}p_i\left(\frac{1}{n}\right)\right] \end{align*}

Finally, we can notice that

$$ \sum\limits_{i=0}^n \left[\frac{1}{i!}p_i\left(\frac{1}{n}\right)\right] $$

is really just some polynomial $q_n \in \mathbb{R}[X]$ evaluated at $\frac{1}{n}$, so we get a final form of:

$$ e = \lim_{n \rightarrow \infty} \sum\limits_{i=0}^n \frac{1}{i!} + \lim_{n \rightarrow \infty} \frac{1}{n} q_n\left(\frac{1}{n}\right)$$

Now, I'm not exactly sure where to go from here, at least not rigorously. That right limit is clearly $0$, but I lack the analysis knowledge to be able to actually prove it. The thing is though, with the $\frac{1}{i!}$ additional coefficient to each of those polynomials, and the fact that in the limit,

$$ p_i\left(\frac{1}{n}\right) \rightarrow p_i(0)=\frac{i(i-1)}{2} $$

(using the sum of the first $i$ integers formula, and it's pretty easy to see that the constant term of $p_i$ is the sum of the first $i$ integers), the product of the two is a smaller and smaller constant with each $i$, the $\frac{1}{n}$ factor clearly "wins", and the limit is $0$. Though it is clearly cheating, that sum must be $0$, because the left limit is the limit you're trying to prove (which you know is correct), so I know that limit must be $0$. Nevertheless, if you believe all that hand-waving about proving it's $0$, then you get:

$$ e = \lim_{n \rightarrow \infty} \sum_{i=0}^n \frac{1}{i!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots $$

Answer 2

$$\prod_{k=1}^n\left(1-\frac{k-1}{n}\right)=\prod_{k=1}^n\frac{1}{n}\left(n-k+1\right)=\frac{1}{n^n}\big[n(n-1)\cdots(1)\big]=\frac{n!}{n^n}$$ Hence, we have: $$\lim_{n\to\infty}\frac{1}{n!}\prod_{k=1}^n\left(1-\frac{k-1}{n}\right)=\lim_{n\to\infty}\frac{n!}{n^n\cdot n!}=\lim_{n\to\infty}\frac{1}{n^n}=0$$ This should be expected, as if the last term approached any other value then the terms of the summation would not approach zero.