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I happened to read a statement related to my question in Ferguson's A Course in Large Sample Theory, where he stated:

$$\left(1 + \frac{\lambda_n(e^{it} - 1)}{n}\right)^n \to \exp[\lambda(e^{it} - 1)],$$ as $n \to \infty$, where $\lambda_n \to \lambda$ as $n \to \infty$.

So it looks like he used the property

$$\left(1 + \frac{z}{n}\right)^n \to e^z \tag{*}$$

as $n \to \infty$ tacitly, even when $z$ is a complex variable.

I understand that under the real setting, the proof of $e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n$ relies on the ordering property of $\mathbb{R}$, but it is unlikely the same technique can be transferred to the complex domain.

Nevertheless, I guess this relation $(*)$ should be correct, but how do we show it rigorously (probably by using some complex analysis tools)?

Zhanxiong
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  • What happens when you split everything into real and imaginary parts? – Quinn Culver Dec 09 '15 at 15:53
  • @ByronSchmuland Indeed. There's a very simple proof of the result that nobody seems to have noticed in that other thread. I just added an answer there - was that the Right Thing to do? – David C. Ullrich Dec 09 '15 at 16:12
  • Someone pointed out this is a duplicate of http://math.stackexchange.com/questions/374747/if-z-n-to-z-then-1z-n-nn-to-ez/1567761#1567761 and then deleted his comment. It is a duplicate of that question. A new answer to that other question has just appeared, imo simpler than the others... – David C. Ullrich Dec 09 '15 at 16:26

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I want to stress that this is an incomplete idea to go about this.

Assume that the function

$f(z)= \lim_{n \to \infty} \Big(1 + \frac{z}{n}\Big)^n - \exp(z)$

is analytic.

On the real line, it coincides with $0$ because $\lim_{n \to \infty} \Big(1 + \frac{x}{n}\Big)^n = \exp(x)$.

Since analytic functions can have at most countably many zeroes, it follows that $f(z) = 0$ for all $z \in \mathbb{C}$.

Tien Truong
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    Here you assume already that the limit exists for $z \in \Bbb C$, and is an analytic function. – Martin R Dec 09 '15 at 16:04
  • Of course this is a valid argument, provided that you are willing to take for faith that this exists and it is analytic, but by no means it is a complete proof. – Wojowu Dec 09 '15 at 16:08
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    See my answer to http://math.stackexchange.com/questions/374747/if-z-n-to-z-then-1z-n-nn-to-ez/1567761#1567761 for a completed version of this argument. – David C. Ullrich Dec 09 '15 at 16:11
  • @Wojowu and Martin R: Yes, I am aware of that. It is sufficient to show that $g(z) = \lim_{n \to \infty} (1 + z/n)^n$ is analytic. My idea is that we can use the binomial expansion to write $g(z)$ as a polynomial and then consider the limit. – Tien Truong Dec 09 '15 at 16:12
  • It's much simpler than that! See my answer in that other thread... – David C. Ullrich Dec 09 '15 at 16:14
  • ??? Someone appears to have +1-ed my comment here, and then upvoted a different answer in that other thread. Whoever, look at the answer I just posted - you'll like it. – David C. Ullrich Dec 09 '15 at 16:27
  • I upvoted your comment here but I haven't done anything in the other thread because I am still saving your proof for later. :D – Tien Truong Dec 09 '15 at 16:43