Recall that e i θ ( cos θ + i sin θ ) Note that ∣ ∑ n = 1 N e θ n t ∣ = ∣ e i θ ( 1 − e i N θ ) 1 − e i θ ∣ ≤ 2 ∣ 1 − e i θ ∣ < ∞ since e i ≠ 1 Since the sequence { 1 n } n = 1 ∞ is a monotonically decreasing sequence and lim n → ∞ 1 n = 0 . Hence, by the Dirichlet’s test we have that ∑ n = 1 ∞ e i n θ n converges ⟹ s = ∑ n = 1 ∞ e i n θ n = ∑ n = 1 ∞ cos n θ + i sin n θ n converges Consequentially, we have that I m ( s ) = ∑ n = 1 ∞ sin n θ n converges Taking θ = 1 , then we have our desired result. \displaystyle
\text{Recall that $e^{i\theta}(\cos \theta$}+i\sin\theta)\\
\text{Note that}\\
\left|\sum^N_{n=1}e^{\theta n t}\right|= \left|\frac{e^{i\theta}(1-e^{iN\theta})}{1-e^{i\theta}}\right|\leq \frac{2}{|1-e^{i\theta}|}<\infty\text{ since $e^i \neq 1$ }\\
\text{Since the sequence $\{\frac{1}{n}\}^\infty_{n=1}$ is a monotonically decreasing sequence}\\
\text{and $\displaystyle\lim_{n\to \infty}\frac{1}{n}=0$. Hence, by the Dirichlet's test we have that }\\
\sum^\infty _{n=1}\frac{e^{in\theta}}{n} \text{ converges}\\
\implies s = \sum^\infty _{n=1}\frac{e^{in\theta}}{n} = \sum^\infty _{n=1}\frac{\cos{n\theta}+i\sin{n\theta}}{n} \text{ converges}\\
\text{Consequentially, we have that}\\
Im(s) = \sum^\infty_{n=1}\frac{\sin{n\theta}}{n} \text{ converges}\\
\text{Taking $\theta = 1$, then we have our desired result.} Recall that e i θ ( cos θ + i sin θ ) Note that ∣ ∣ n = 1 ∑ N e θ n t ∣ ∣ = ∣ ∣ 1 − e i θ e i θ ( 1 − e i Nθ ) ∣ ∣ ≤ ∣1 − e i θ ∣ 2 < ∞ since e i = 1 Since the sequence { n 1 } n = 1 ∞ is a monotonically decreasing sequence and n → ∞ lim n 1 = 0. Hence, by the Dirichlet’s test we have that n = 1 ∑ ∞ n e in θ converges ⟹ s = n = 1 ∑ ∞ n e in θ = n = 1 ∑ ∞ n cos n θ + i sin n θ converges Consequentially, we have that I m ( s ) = n = 1 ∑ ∞ n sin n θ converges Taking θ = 1, then we have our desired result.
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