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I've had no luck with this one. None of the convergence tests pop into mind.

I tried looking at it in this form $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ and apply Dirichlets test. I know that $\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n} \to 0$ but not sure if it's decreasing.

Regarding absolute convergence, I tried:

$$|\sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}|\geq \sin^2 n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}=$$

$$=\frac{1}{2}\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}-\frac{1}{2}\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$$

But again I'm stuck with $\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$.
Assuming it converges then I've shown that $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ doesn't converge absolutely.

Quixotic
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daniel.jackson
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  • Perhaps there's something useful to be done with the fact that $\log n<1+\frac{1}{2}+\cdots+\frac{1}{n}<\log n+1$? – Isaac Dec 31 '10 at 11:14
  • Also, Mathematica seems to think that it does in fact converge, giving a result of 1/2 I E^-I (E^I EulerGamma Log[1-E^I]-E^I EulerGamma Log[(-1+E^I) E^-I]-(Hypergeometric2F1Regularized^(0,0,1,0))[1,1,2,E^-I]+E^(2 I) (Hypergeometric2F1Regularized^(0,0,1,0))[1,1,2,E^I]), which is approximately $1.05895$. – Isaac Dec 31 '10 at 11:19
  • Just a comment, $$\sum_{n=1}^{\infty} \frac{\sin n}{n} = \frac{\pi}{2} - \frac{1}{2}$$Maybe this might help... – Roupam Ghosh Dec 31 '10 at 11:38
  • My older copy (5.2) of Mathematica yields the result $$\frac{i}{2}\left(\mathrm{Li}_2\left(1+\frac1{\exp(i)-1}\right)-\mathrm{Li}_2\left(\frac1{1-\exp(i)}\right)\right)$$ where $\mathrm{Li}_2(x)$ is the dilogarithm. Numerically, it agrees with @Isaac's result. I am away from my refs so I cannot analytically prove the equivalence of the two. – J. M. ain't a mathematician Dec 31 '10 at 11:53

2 Answers2

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The sequence $$a_n=\frac{1+\frac{1}{2}+\cdots +\frac{1}{n}}{n}$$ is decreasing because proving $a_{n+1}<a_n$ reduces to proving $$\frac{n}{n+1}<1+\frac{1}{2}+\cdots +\frac{1}{n}$$ which is clearly true.

Also the sums $$s_k=\sum_{n=1}^k \sin n$$ are bounded. (This can be easily proved by writing $\sin n$ in its complex form and using finite geometric formula; in fact $|s_k|$ are bounded by $\frac{1}{|1-e^i|}+\frac{1}{|1-e^{-i}|}$).

Furthermore $a_n\to 0$ as $n\to \infty$ because $1+\frac{1}{2}+\cdots +\frac{1}{n}\approx \log n$.

So the series converges by Dirichlet test. As Shai shows below, the convergence is not absolute.

TCL
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14

To show that the series does not converge absolutely, it suffices to show that $\sum\nolimits_{n = 1}^\infty {|\frac{{\sin n}}{n}|} = \infty $. For this purpose, see this: the paragraph starting with "It's not absolutely convergent" as well as the one starting with "This series exhibits rather irregular behaviour" (two different approaches).

Shai Covo
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  • How does one show the two inequalities needed on the page you referenced? $\min(|\sin(n)|,|\sin(n+1)|) \geq m > 0$ and $|\sin(x)| + |\sin(x-1)| \geq \sin(1)$ – admchrch Mar 26 '11 at 22:00
  • @admchrch: thanks for pointing this out. – Shai Covo Mar 27 '11 at 20:58
  • @admchrch: For the first, there should have been $\max$ instead of $\min$; for the second, at least it can be confirmed by plotting the difference $|\sin(x)| + |\sin(x-1)| - \sin(1)$. – Shai Covo Mar 27 '11 at 21:25
  • The method that uses the first bound is invalid if it is max instead of min. I guess the second can be proved by considering the functions $\sin(x) + \sin(x-1) - \sin(1)$, $\sin(x) - \sin(x-1) - \sin(1)$, $-\sin(x) + \sin(x-1) - \sin(1)$, and $-\sin(x) - \sin(x-1) - \sin(1)$ in appropriate regions and using calculus. A more elegant method of proof does not occur to me. – admchrch Mar 28 '11 at 03:48
  • To disprove absolute convergence, I noted $|\sin n| \geq \sin^2 n = \frac{1}{2}( 1 - \cos 2n )$ and used Dirichlet's test to show $\sum \frac{\cos 2n}{n}$ diverges. – admchrch Mar 28 '11 at 03:54
  • I suspect it is true that $|\sin n| \geq m > 0$, but I think the proof requires some nontrivial number theory. – admchrch Mar 28 '11 at 04:07
  • @admchrch: For the first method, what's wrong if it is $\max$ instead of $\min$? – Shai Covo Mar 28 '11 at 07:08
  • @Shai: Perhaps I am missing something. Doesn't the first method work by asserting $$\frac{|\sin n|}{n} \geq \frac{\min(|\sin n|, |\sin(n-1)|)}{n} \geq \frac{m}{n}$$ and deducing divergence by comparison with the harmonic series? The first inequality is reversed if it is max instead of min. – admchrch Mar 28 '11 at 07:15
  • @admchrch: The first method works thanks to $\frac{{|\sin n|}}{n} + \frac{{|\sin (n + 1)|}}{{n + 1}} \ge \frac{{|\sin n| + |\sin (n + 1)|}}{{n + 1}} \ge \frac{m}{{n + 1}}$. – Shai Covo Mar 28 '11 at 07:49
  • @admchrch: Concerning a previous comment, $\lim \inf _{n \to \infty } |\sin n| = 0$. – Shai Covo Mar 28 '11 at 07:51
  • @Shai: You are right, it is not true that $|\sin n|≥m>0$. – admchrch Mar 28 '11 at 07:58
  • @Shai: Okay I see it now: $$|\sin(n)| + |\sin(n+1)| \geq \max(|\sin(n)|,|\sin(n+1)|)$$ Thanks. – admchrch Mar 28 '11 at 08:01
  • This also follows easily from a Dirichlet approximation type Theorem. More exactly, because $2\pi$ is irrational, one can show that there exists an $N$ so that among any $N$ consecutive integers we can find an $n$ with $\sin(n) \geq \frac{1}{2}$. – N. S. Nov 07 '11 at 03:33
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    Link's down.... – Gabriel Romon Jun 21 '18 at 05:37