Answer to Question #203265 in Real Analysis for Raj kumar

Question

Test the series:

_n=1 ∑∞ (-1)^n-1 [Sin(nx)]/n√n

for absolute and conditional convergence.

Answer 1

$\displaystyle\sum_{n=1}^{\infin}\dfrac{(-1)^n\sin[nx]}{n\sqrt{n}}$

$|\sin[nx]|\leq1, x\in\R, n\geq1$

Then

$\bigg|\dfrac{(-1)^n\sin[nx]}{n\sqrt{n}}\bigg|\leq\dfrac{1}{n\sqrt{n}}, x\in\R, n\geq1$

The series $\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n\sqrt{n}}$ converges as $p$ -series with $p=\dfrac{3}{2}>1.$

Therefore the series $\displaystyle\sum_{n=1}^{\infin}\dfrac{(-1)^n\sin[nx]}{n\sqrt{n}}$ converges absolutely by the Comparison Test, for $x\in\R.$