For this question, I can successfully prove that the series of |sin(n^0.5)|/n^1.5 converges but I don't understand why the solution can just say " |sin(n^0.5)|/n^1.5 converges, hence the original series (-1)^n sin(n^0.5)|/n^1.5 converges absolutely." Can someone please explain me why the solution can say the negative case converges straight away but no need to use the Alternative Series Test?
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The alternating series test is actually not applicable here because one has no idea what the sign of $\sin(\sqrt{n})$ is. However, the solution is just to use the comparison test. This eliminates both the $\sin$ factor and the sign factor. Then, one uses the integral test or the dyadic summation test together with geometric series if you like to verify that $\sum_{n\ge1} 1/{n^{3/2}}$ converges to a finite value.
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You may just write that $$ \left|\sum_{n=1}^{\infty}(-1)^n \frac{\sin(\sqrt{n})}{n^{3/2}}\right|\leq \sum_{n=1}^{\infty}\left|\frac{\sin(\sqrt{n})}{n^{3/2}}\right|\leq \sum_{n=1}^{\infty}\frac{1}{n^{3/2}} $$ and your initial series is absolutely convergent thus is convergent, where we have used the fact that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{3/2}} $ is convergent.
If a series converges absolutely then it's a convergent series. The proof is based on the Cauchy criterion for the convergence of the series.