Test the convergences of the following series
$$ \sum_{n=1}^{\infty} (-1)^n \frac{ \sin n}{\sqrt{n}}$$
Asked
Active
Viewed 74 times
5
-
I suspect it converges, but probably the signs of the sines behave in a somewhat complicated way. – Michael Hardy Mar 27 '13 at 14:52
-
You might want to use Abel's summation formula : http://en.wikipedia.org/wiki/Abel's_summation_formula – Joel Cohen Mar 27 '13 at 15:09
-
$\sum \frac{e^{i\pi n +in}}{\sqrt{n}}$ – Sungjin Kim Mar 27 '13 at 15:37
1 Answers
0
Denote $\theta = 1 + \pi$, $a_n = (-1)^n \sin n = \sin(n \theta)$ and $\phi(x) = \frac{1}{\sqrt{x}}$. Then compute
$$A(x) = \sum_{1 \le k \le x} a_k = \textrm{Im}\left(\sum_{1 \le k \le x} e^{i k \theta}\right) = \textrm{Im}\left(\frac{e^{i (\lfloor x \rfloor +1) \theta}-e^{i \theta}}{e^{i \theta}-1}\right)$$ Which yields $$|A(x)| \le \left|\frac{e^{i (\lfloor x \rfloor+1) \theta}-e^{i \theta}}{e^{i \theta}-1}\right| \le \frac{2}{|e^{i \theta}-1|}$$
So $x \mapsto A(x)$ is bounded. Using Abel's summation formula, you may write
$$\sum_{1 \le n \le x} a_n \phi(n) = A(x)\phi(x) - \int_1^x A(x)\phi'(x) \, dx$$
Since $A$ is bounded, and $\phi$ goes to zero at infinity, you may check that both terms in the right hand side of the above expression have a limit at infinity.
Joel Cohen
- 9,289