Prove that the sequence \(\left(1+\frac{z}{n}\right)^{n}\) converges uniformly to \(e^{z}\) on compact subsets of \(\mathbb{C}\).
Hint: \(e^{n \log w_{n}}=w_{n}^{n}\) and \(e^{z}\) is uniform continuous on compact subsets of \(\mathbb{C}\).
Solution
Let \(K\) be compact, where \(z\in K\implies \abs{z} \leq R\) for some constant \(R\). For the remainder of the problem, we only work in \(K\).
\(f_n(z) \da n\log(1 + {z\over n}) \to z\) uniformly.
\(f_n\) are uniformly bounded on \(K\).
\(e^z\) is uniformly continuous on \(K\).
If \(g_n\to g\) uniformly and \(F\) is uniformly continuous, then \(F \circ g_n \to F\circ g\) uniformly.
Why these claims imply the result:
If \(f_n(z)\to z\) uniformly, both are uniformly bounded, and \(e^z\) is uniformly continuous, then \(e^{f(z)}\to e^z\) uniformly.
(Of first claim)
We’ll first show that for \(w\) in a neighborhood of zero avoiding 1, there exists a constant \(C\) such that \[ \abs{ 1 - {\log(1+w) \over w} } \leq C\abs{w} .\] This follows from estimating the series expansion about \(w=0\): \[ \abs{ 1 - {\log(1+w) \over w} } &= \abs{\inverseof{w}\sum_{k\geq 1} { (-w)^k \over k} } \\ &= \abs{\sum_{k\geq 2} {(-w)^{k-1} \over k} } \\ &\leq {\sum_{k\geq 2} {\abs{w}^{k-1} \over k} } \\ &= {\sum_{k\geq 1} {\abs{w}^{k} \over k+1} } \\ &\leq {\sum_{k\geq 1} {\abs{w}^{k} \over 2} } \\ &= {1\over 2}\qty{{1\over 1 - \abs w} - 1 } \\ &= {1\over 2}\abs{2} \qty{1\over 1 - \abs w} \\ &\leq C \abs{w} ,\] using that \({1\over 1-x}\) is bounded in compact sets avoiding \(x=1\).
We can now apply the \(M\dash\)test: \[ \abs{n\log\qty{ 1 + {z\over n} } - z } &= \abs{z}\cdot \abs{ {{ \log\qty{1 + {z\over n}} \over {z\over n}} - 1} } \\ &\leq \abs{z} \cdot C\abs{z\over n} \\ &\leq M\cdot C\qty{M\over n} \\ &= {CM^2 \over n}\\ &\convergesto{n\to\infty}0 .\]