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Let $f_n(z) = (1-z^2/n)^n$, and let $f(z)=\operatorname{exp}(-z^2)$. I need to show that $f_n$ converges uniformly to $f$ in any closed disc.

I saw this: Uniform Convergence of an Exponential Sequence of Functions, but I'm not sure the methods used there are applicable here because we're talking about the complex plane, and the logarithm function doesn't behave nicely there.

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Why can't you use Peter's answer? (no $\log$ involved) –  draks ... May 10 '12 at 12:40
    
Thanks. Must've missed that. I'm trying to expand e^z to degree n such as the remainder is epsilon/2, and this is what I get: $\left|e^w-(1+w/n)^n\right| = \left|\sum\limits_{k=0}^n \frac{w^k}{k!} + w^{n+1}g_{n+1}(w) - \sum\limits_{k=0}^n \binom{n}{k} \frac{w^k}{n^k}\right| \leq \frac{\varepsilon}{2} + \sum_{k=0}^n \frac{|w|^k}{k!}\left|\frac{n!}{(n-k)!n^k}-1\right|$ I can't really continue from here because $|w|^k$ can be quite large as $k \approx n \to \infty$. –  George May 10 '12 at 16:15

1 Answer 1

Don't be afraid to use the logarithm in the complex plane. Given a closed disk $\{z:|z|\le R\}$, consider only $n>R^2$: this guarantees $|z^2/n|<1$, and therefore places $1-z^2/n$ in the open right halfplane, where we have the principal branch of $\log$. By the definition of derivative, $$\lim_{n\to\infty } n\log(1-z^2/n)= -z^2 \lim_{n\to\infty } \frac{\log(1-z^2/n)-\log 1}{-z^2/n} = -z^2$$ (using $(\log \zeta)'=1$ at $\zeta=1$). This is a lot easier than messing with binomial coefficients.

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