Let $x_n\to x$ be a convergent sequence of real numbers with limit $x\in\mathbb R$.
Then
$$\lim_{n\to\infty}\prod_{k=1}^n \left(1+\frac{x_k}{n}\right)=e^x.$$
Is this true?
(Note: The formula has $n$ in the denominator and not the index $k$.)
Here is an attempt at a proof:
$$\begin{aligned}
\prod_{k=1}^n \left(1+\frac{x_k}{n}\right)&=\prod_{k=1}^n \left(1+\frac xn+\frac{x_k-x}{n}\right)\\
&=\left(1+\frac xn\right)^n+\left(1+\frac xn\right) \prod_{k=2}^n \left(\frac{x_k-x}{n}\right) +\text{stuff}\\
\end{aligned}$$
where the "stuff" will include many terms that look like a power of $\left(1+\frac xn\right)$ times some product of factors like $\left(\frac{x_k-x}{n}\right)$ and thus every other term will be decaying to zero (as $n\to\infty$) no matter how slow $x_n$ converges to $x$.
Thus taking the limit as $n\to\infty$ gives the desired result.
Now the problem is that we have infinite products still in there like $\prod_{\text{some } k} \left(\frac{x_k-x}{n}\right)$, but each factor in the infinite product has form $\frac{x_k-x}{n}$ and hence goes to zero. But I am very uncertain about this reasoning. I worry the infinite products might be very badly behaved.
I'm just wondering what technical issues I might be overlooking, and if I am making any mistakes in my argument, or if the argument is sort of correct but insufficient according to common practice.
Best Answer
I'm not convinced. The "stuff" consists of various small terms that all will become very small, but at the same time, there will be more and more of these terms as $n \to \infty$. As with a Riemann sum, we expect to sum a lot of small quantities and take the limit, but this doesn't imply the Riemann integral is always $0$.
Here's how I would approach it:
Fix $\varepsilon > 0$, and let $N \in \Bbb{N}$ be such that $$n \ge N \implies |x_n - x| < \varepsilon$$ and $x_n > -N$ (which we can do, as convergent sequences are bounded). Then, for $n > N$, $$\prod_{k=1}^n \left(1 + \frac{x_k}{n}\right) = \prod_{k=1}^{N-1} \left(1 + \frac{x_k}{n}\right) \cdot \prod_{k=N}^{n} \left(1 + \frac{x_k}{n}\right),$$ where $$\left(1 + \frac{x - \varepsilon}{n}\right)^{n-N+1} \le \prod_{k=N}^{n} \left(1 + \frac{x_k}{n}\right) \le \left(1 + \frac{x + \varepsilon}{n}\right)^{n-N+1}.$$ Since $x_n > -N$ and $n > N$, we have $$\prod_{k=1}^{N-1} \left(1 + \frac{x_k}{n}\right) > 0,$$ so $$\left(1 + \frac{x - \varepsilon}{n}\right)^{n-N+1} \prod_{k=1}^{N-1} \left(1 + \frac{x_n}{n}\right) \le \prod_{k=1}^{n} \left(1 + \frac{x_k}{n}\right) \le \left(1 + \frac{x + \varepsilon}{n}\right)^{n-N+1} \prod_{k=1}^{N-1} \left(1 + \frac{x_n}{n}\right).$$ Taking the limit as $n \to \infty$, $$e^{x - \varepsilon} \le \prod_{k=1}^{\infty} \left(1 + \frac{x_k}{n}\right) \le e^{x + \varepsilon},$$ for all $\varepsilon > 0$. So, if you accept that the product converges, the limit must be $e^x$, by the continuity of $e^x$.