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Using Taylor's theorem applied to a branch of log $(1 + z/n),$ prove that lim$_{n \rightarrow \infty} (1 + \frac{z}{n})^{n} = e^z$ uniformly on all compact sets.

Taylor's theorem gives us that $$f(z) = f(a) + \frac{f'(a)}{1!}(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \cdots+\frac{f^{(m-1)}}{(m-1)!}(z-a)^{m-1} + f_m(z)(z-a)^m.$$

If $f(z) =$ log$(1 + z/n)$, then $$f'(z) = \frac{1}{1+z/n}\cdot\frac{1}{n},$$ $$f''(z) = -\frac{1}{(1 + z/n)^2}\cdot\frac{1}{n^2}, $$ $$f^m(z) = \frac{(m-1)!(-1)^{m-1}}{(1 + z/n)^m}\cdot\frac{1}{n^m}.$$

My issue is that I don't see how this is all going to come together to prove what I need to.

Jerry
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1 Answers1

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Some useful facts here: 1. Suppose $f_1,f_2,\dots$ are complex valued functions on a set $E,$ and $f_n \to f$ uniformly on $E.$ If $f$ is bounded on $E,$ then $e^{f_n} \to e^f$ uniformly on $E.$ 2. Because $\log'(1)=1,$ $n\log(1+z/n) \to z$ uniformly on compact subsets of $\mathbb C.$

Use these results to see $\exp (n\log(1+z/n)) \to e^z$ uniformly on compact subsets of $\mathbb C.$ But $\exp (n\log(1+z/n)) = (1+z/n)^n,$ so we're done.

zhw.
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