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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.

  • 0
    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.2010-12-31

2 Answers 2

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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} 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.

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    Oh yeah. Thanks. =)2011-11-13
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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).

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    Link's down....2018-06-21