Prove that the following series converges uniformly on the set \(\theset{z \suchthat \Im(z) < \ln 2}\): \[ \sum_{n=1}^\infty {\sin(nz) \over 2^n} .\]
Solution
The statement is false on the stated half-plane. If \(z=-iy\) with \(y>\log2\), then \[ \left|\frac{\sin(nz)}{2^n}\right| =\frac{\sinh(ny)}{2^n} \sim \frac12\left(\frac{e^y}{2}\right)^n, \] so even the terms fail to tend to zero.
The exact pointwise convergence region is the strip \[ |\Im z|<\log2. \] Indeed, writing \(z=x+iy\), \[ |\sin(nz)| \le \frac12\left(e^{n|y|}+e^{-n|y|}\right) \le e^{n|y|}. \] Hence if \(|y|<\log2\), then \[ \frac{|\sin(nz)|}{2^n}\le \left(\frac{e^{|y|}}2\right)^n, \] which is summable. Conversely, if \(|y|>\log2\), the terms do not tend to zero by the same exponential asymptotics; at \(|y|=\log2\) they also fail to tend to zero for suitable \(x\).
Moreover the convergence is uniform on every closed substrip \(|\Im z|\le a<\log2\), by the Weierstrass \(M\)-test with majorant \((e^a/2)^n\). It is not uniform on the whole open strip because the suprema of the individual terms there do not tend to zero.