Answer to Question #262920 in Real Analysis for Danny

Question #262920

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



1
Expert's answer
2021-11-15T02:21:48-0500

Let us show that the series "\\sum\\limits_{n=1}^{\\infty}\\frac{\\sin(n)}n" is divergent.

Note that "|\\sin x|\u2a7e\\frac{1}2" for "x\u2208I_k=[\\frac{\u03c0}6+k\u03c0,\u03c0\u2212\\frac{\u03c0}6+k\u03c0]."

Let us find the length of every "I_k:"

"|I_k|=\u03c0\u2212\\frac{2\u03c0}6=\\frac{2\u03c0}3>2."

We conclude that every "I_k" contains at least one natural number "n_k."

Then

"\\sum\\limits_{n=1}^N\\frac{|\\sin(n)|}n\n\u2a7e\\frac{1}2\\sum\\limits_{n_k\\le N}\\frac{1}{n_k}\\to\\infty" when "N\u2192\u221e"

since the harmonic series "\\sum\\limits_{n=1}^{\u221e}\\frac{1}n" diverges.


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