For every real number \(c\) with \(0\leq c<\ln 2\), prove that the series \[ \sum_{n=1}^\infty \frac{\sin(nz)}{2^n} \] converges uniformly on \[ \{z:\abs{\operatorname{Im}z}\leq c\}. \] Deduce that the series converges locally uniformly on \[ \{z:\abs{\operatorname{Im}z}<\ln2\}. \]
Erratum: the statement is false as written in the source. For \(z = iy\) we have \(\abs{\sin(nz)} = \sinh(n\abs{y})\), so at \(z = -2i\) the terms have modulus \(\sinh(2n)/2^n \to \infty\) and the series diverges, although \(\operatorname{Im}(-2i) < \ln 2\). Even on the strip \(\{\abs{\operatorname{Im} z} < \ln 2\}\), where the series converges, the convergence is not uniform: at \(z = iy\) with \(0 < y < \ln 2\) the terms are \(i\sinh(ny)/2^n\), with \(\sinh(ny)/2^n \ge (e^{y}/2)^n/4\) for \(ny \ge 1\), and \(e^y/2 \to 1\) as \(y \to \ln 2\), so no tail is uniformly small. The correct statement is that the series converges uniformly on \(\{z : \abs{\operatorname{Im} z} \le c\}\) for each \(0\le c < \ln 2\), hence locally uniformly on \(\{z : \abs{\operatorname{Im} z} < \ln 2\}\).
Solution
Fix \(c\) with \[ 0\leq c<\ln2 \] and set \[ q=\frac{e^c}{2}. \] Then \(0<q<1\).
1. If \(\abs{\operatorname{Im}z}\leq c\), then for every integer\(n\geq1\),\[ \abs{\sin(nz)}\leq e^{nc}. \]
Write \(z=x+iy\). Using \[ \sin(nz) = \frac{e^{inz}-e^{-inz}}{2i}, \] one has \[ \abs{e^{inz}}=e^{-ny}, \qquad \abs{e^{-inz}}=e^{ny}. \] Hence \[ \begin{aligned} \abs{\sin(nz)} &\leq \frac{e^{-ny}+e^{ny}}2\\ &= \cosh(ny)\\ &\leq e^{n\abs{y}}\\ &\leq e^{nc}. \end{aligned} \]
2. The series converges uniformly on\[ \{z:\abs{\operatorname{Im}z}\leq c\}. \]
By step 1, \[ \abs{\frac{\sin(nz)}{2^n}} \leq \left(\frac{e^c}{2}\right)^n = q^n \] throughout the closed strip. Since \(0<q<1\), the numerical geometric series \[ \sum_{n=1}^{\infty}q^n \] converges. The Weierstrass M-test therefore gives uniform convergence on the closed strip.
3. The series converges locally uniformly on\[ S=\{z:\abs{\operatorname{Im}z}<\ln2\}. \]
Let \(K\subseteq S\) be compact. The continuous function \(z\mapsto\abs{\operatorname{Im}z}\) attains a maximum \(c_K\) on \(K\). Because \(K\subseteq S\), \[ c_K<\ln2. \] Set \[ c=\frac{c_K+\ln2}{2}. \] Then \[ c_K<c<\ln2, \] and \[ K\subseteq\{z:\abs{\operatorname{Im}z}\leq c\}, \] so step 2 gives uniform convergence on the larger closed strip, hence on \(K\).
4. Q.E.D.
Step 2 proves the corrected uniform-convergence statement for every closed substrip, and step 3 gives the stated local uniform convergence.