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It is a well known fact that

$$\lim_n (1+x/n+O(n^{-3/2}))^n=e^x$$

For example, this is a key step in the standard proof of the central limit theorem.

What can we say about the rate of convergence of this approximation?

More precisely, suppose we expand out the first term of the remainder:

$$(1+x/n+y/n^{3/2}+O(1/n^2))^n$$

Is it correct to say that this is equal to $e^x+C/\sqrt{n}+O(n^{-3/2})$? And if so, is there an explicit expression for $C$ in terms of x and y?

Simon Segert
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    $(1+c_n/n)^n\xrightarrow{n\rightarrow\infty}e^c$ whenever $c_n$ is a sequence of Complex numbers converging to $c$. This has been discussed many times in MSE, for example here – Mittens Jun 03 '24 at 21:21
  • in your case, take $c_n=x+\frac{y}{n^{1/2}}+O(\tfrac1n)$. – Mittens Jun 03 '24 at 21:26
  • @Mittens, thanks but that does not address my question which is "what can we say about the rate of convergence?" – Simon Segert Jun 03 '24 at 22:08

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$\ln ((1+x/n)^n)=n\ln(1+x/n)=n(x/n-x^2/(2n^2)+...)$ so $(1+x/n)^n=e^{x-x^2/(2n)+...}=e^x(1-\frac{x^2}{2n}+O(x^3/n^2))$

van der Wolf
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