# Does the sum $\sum _{n=1}^{\infty }\left(\frac{\sin\left(n\right)}{n}\sum _{k=1}^n\left(\frac{1}{k}\right)\:\right)$ absolutely converge?

I have the following sum:

$$\sum _{n=1}^{\infty }\left(\frac{\sin\left(n\right)}{n}\sum _{k=1}^n\left(\frac{1}{k}\right)\:\right)$$

And I need to determine if it absolutely converges, conditionally converges or diverges.

I know it converges conditionally with the help of the Dirichlet test, but I'm having a hard time determining if it converges absolutely. Intuitively, I have a feeling it does not, because $\left|\frac{\sin\left(n\right)}{n}\right|$ looks to be partial to the harmonic sum, which diverges.

• Do you know that $\sum\frac{\sin n}{n}$ isn't absolutely convergent? – user63181 May 28 '14 at 12:15
• – gammatester May 28 '14 at 12:35

$$\sum _{n=1}^{\infty}\left|\frac{\sin\left(n\right)}{n}\sum _{k=1}^n\left(\frac{1}{k}\right)\:\right| \ge \sum _{n=1}^{\infty }\left|\frac{\sin\left(n\right)}{n}\:\right|$$ We know that the right side does not converge hence the series does not converge.
• @Alex $|\sin(x)|\ge \sin^2(x)$ using this inequality, you get $$\frac{|\sin (n)|}{n} \ge \frac{\sin^2 (n)}{n} = \frac{1 - \cos 2n}{2n}$$ – Santosh Linkha May 28 '14 at 12:52
• @Alex $\frac{\cos 2n}{2n}$ converges by the same test that works for $\frac{\sin n }{n}$ i.e. Dirichlet Test, while the $\frac 1 n$ does not converge by integral test, Hence it does not converge by comparison test. – Santosh Linkha May 28 '14 at 12:54