# Does $\sum \limits_{n=1}^\infty\frac{\sin n}{n}(1+\frac{1}{2}+\cdots+\frac{1}{n})$ converge (absolutely)?

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.

• Perhaps there's something useful to be done with the fact that $\log n<1+\frac{1}{2}+\cdots+\frac{1}{n}<\log n+1$? 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 $1.05895$. 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... 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. Dec 31 '10 at 11:53

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.

• Are you sure about that first claim? I'm convinced that sequence is indeed decreasing, but I don't see how it reduces to the inequality you have there. Nov 13 '11 at 16:24
• Just write out $a_{n+1}<a_n$ and multiply both sides by $n(n+1)$.
– TCL
Nov 13 '11 at 16:32
• Oh yeah. Thanks. =) Nov 13 '11 at 16:44

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).

• How does one show the two inequalities needed on the page you referenced? $\min(|\sin(n)|,|\sin(n+1)|) \geq m > 0$ and $|\sin(x)| + |\sin(x-1)| \geq \sin(1)$ Mar 26 '11 at 22:00
• @admchrch: thanks for pointing this out. 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 $|\sin(x)| + |\sin(x-1)| - \sin(1)$. 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. Mar 28 '11 at 3:48
• Link's down.... Jun 21 '18 at 5:37