How does one determine whether or not the infinite series $$\sum_{n=1}^\infty \sin^n(n)$$ converges? I suspect that it doesn't converge absolutely, but I have no idea how to prove/disprove convergence or prove/disprove absolute convergence. Help?
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Eric Wofsey
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Franklin Pezzuti Dyer
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2I just feel like sharing this wonderful, somewhat related solution/problem: https://mathoverflow.net/questions/282259/is-the-series-sum-n-sin-nn-n-convergent/282290#282290 – mathworker21 Jan 20 '18 at 17:04
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1Does the general term go to zero? – hamam_Abdallah Jan 20 '18 at 17:17
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The answer is negative. Quoting the first part of the answer I gave to this related question,
The irrational number $\frac{\pi}{2}$ has an infinite number of convergents $\frac{p_n}{q_n}$ with an odd denominator, hence $\left|\frac{p_n}{q_n}-\frac{\pi}{2}\right|\leq\frac{1}{q_n^2}$ gives: $$ \left|\sin(p_n)\right| = \left|\sin\left(\frac{\pi}{2}q_n+\frac{\theta}{q_n}\right)\right|=\left|\cos\frac{\theta}{q_n}\right|,\quad |\theta|\leq 1,$$ $$ \left|\sin(p_n)\right|\geq 1-\frac{1}{q_n^2}$$ hence $\left|\sin n\right|^n$ is bigger that $\left(1-\frac{1}{n^2}\right)^n$ infinitely often.
In particular the main term of the given series does not converge to zero, so such series is not convergent (either conditionally or absolutely).
Jack D'Aurizio
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