How the series is convergent and absolute convergent

Question

$$\sum_{n=1}^\infty (-1)^{n-1} \frac{\sin(nx) + \cos(nx)}{n^{3/2}}$$ I tried to prove convergence by Leibnitz theorem but can't prove absolute convergent please give me some hint about this.

Answer 1

$\newcommand{\nf}{\infty}$

If you instead mean

$$\sum_{n=1}^{\nf}\frac{(-1)^{n-1}(\sin(nx)+\cos(nx))}{n^{\frac{3}{2}}}$$

Then we have:

$$\sum_{n=1}^{\nf}\left|\frac{(-1)^{n-1}(\sin(nx)+\cos(nx))}{n^{\frac{3}{2}}}\right|=\sum_{n=1}^{\nf}\left|\frac{\sin(nx)+\cos(nx)}{n^{\frac{3}{2}}}\right|$$$$

$$\leq\sum_{n=1}^{\nf}\left|\frac{\sin(nx)}{n^{\frac{3}{2}}}\right|+\sum_{n=1}^{\nf}\left|\frac{\cos(nx)}{n^{\frac{3}{2}}}\right|\leq \sum_{n=1}^{\nf}\frac{1}{n^{\frac{3}{2}}}+\sum_{n=1}^{\nf}\frac{1}{n^{\frac{3}{2}}}=\sum_{n=1}^{\nf}\frac{2}{n^{\frac{3}{2}}}$$

And this last sum can be easily seen to converge because it is bounded above by $2+\int_1^{\nf}\frac{2}{x^{\frac{3}{2}}}dx$ which is obviously finite.

If, however, your question is as intended, then the series does not converge in general on the real line. Take, for instance, $x=\pi$. We have:

$$\sum_{n=1}^{\nf}\frac{(-1)^{n-1}(\sin(n\pi)+\cos(n\pi))}{\pi^{\frac{3}{2}}}=\sum_{n=1}^{\nf}\frac{(-1)^{n-1}(-1)^n}{\pi^{\frac{3}{2}}}=-\frac{1}{\pi^{\frac{3}{2}}}\sum_{n=1}^{\nf}1$$

which is obviously divergence (this divergence can be easily seen similarly for all multiples of $\pi$).

In fact, this divergence of the series holds more generally, as, for any nonzero $x$ we can find infinitely many multiples of $x$ that are arbitrarily near a multiple of $\pi$. Near enough that we can guarantee that infinitely many terms of $\frac{|\sin(nx)+\cos(nx)|}{x^{\frac{3}{2}}}$ are at least $\frac{1}{2x^{\frac{3}{2}}}$. Then, since for any series, $(a_n)_{n\in\mathbb N}$, s.t. $\lim_{n\rightarrow \nf} |a_n|\neq 0$, $\sum_{n=1}^{\nf} a_n$ diverges, this series must diverge as well.