Answer to Question #263433 in Calculus for Sia

Question #263433

Convergence test for "\\displaystyle\\sum_{n=1}^\\infty \\frac{sin(n)}{n}".


1
Expert's answer
2021-11-15T20:22:59-0500

"\\displaystyle\n\\text{Recall that $e^{i\\theta}(\\cos \\theta$}+i\\sin\\theta)\\\\\n\\text{Note that}\\\\\n\\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 $e^i \\neq 1$ }\\\\\n\\text{Since the sequence $\\{\\frac{1}{n}\\}^\\infty_{n=1}$ is a monotonically decreasing sequence}\\\\\n\\text{and $\\displaystyle\\lim_{n\\to \\infty}\\frac{1}{n}=0$. Hence, by the Dirichlet's test we have that }\\\\\n\\sum^\\infty _{n=1}\\frac{e^{in\\theta}}{n} \\text{ converges}\\\\\n\\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}\\\\\n\\text{Consequentially, we have that}\\\\\nIm(s) = \\sum^\\infty_{n=1}\\frac{\\sin{n\\theta}}{n} \\text{ converges}\\\\\n\\text{Taking $\\theta = 1$, then we have our desired result.}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
APPROVED BY CLIENTS