Show that \(\sum_{k\geq 0}z^k/k!\) converges locally uniformly to \(e^z\).
Solution
Let \(K\) be compact, and choose \(M\) with \(\abs{z} \leq M\) for \(z\in K\). Then, with \(\norm{\cdot}_\infty\) the supremum over \(K\), \[ \norm{e^z - \sum_{0\leq k \leq n} z^k/k!}_\infty &= \norm{\sum_{k\geq n+1}z^k/k! }_{\infty} \\ &\leq \sum_{k\geq n+1} M^k /k! .\] The right side is the tail of the convergent series \(\sum_{k\geq0}M^k/k!=e^M\), so it tends to \(0\) as \(n\to\infty\). Hence the partial sums converge to \(e^z\) uniformly on \(K\) (this is the Weierstrass \(M\)-test with \(M_k=M^k/k!\)).