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$$\sum_{n=1}^\infty \cos(n\pi)\sin\left(\frac{\pi}{\sqrt {n}}\right)$$

I know that to show that a series is conditionally convergent, I will have to show that the series is convergent, and also show that the absolute value of the series is divergent.

I am able to show that $$\sum_{n=1}^\infty \cos(n\pi)\sin\left(\frac{\pi}{ \sqrt n}\right)$$ is convergent using the alternating series test, where the limit is equals to 0 and the expression is non-increasing. But I have no idea how to show that the absolute value of the series is divergent. Please correct me if my concept is wrong.

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  • $\begingroup$ Hint: for large $n$, $\sin(\pi/\sqrt n)\approx\pi/\sqrt n$. $\endgroup$ – Yves Daoust Oct 11 '17 at 8:25
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you can use Dirichlet's test: $sin(\frac{\pi}{\sqrt n}) $ is monotonic and $lim_{n\to \infty} sin(\frac{\pi}{\sqrt n}) = 0$ also we shell see that $\forall m > 1 \exists M >= 0 :| \sum_{n=1}^{m} cos(n\pi) | < M$

we know that : $| \sum_{n=1}^{m} cos(n\pi) | (*) <= |\frac{1}{2} ((-1)^m-1)| <= 3 = M $

put notice that we can infer (*) from the integral test (or any other) - it is pretty simple i believe you can do it yourself

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When $n\geq4$ then ${\pi\over\sqrt{n}}\leq{\pi\over2}$, hence $n\mapsto \sin{\pi\over\sqrt{n}}>0$ is monotonically decreasing to $0$ when $n\geq4$. As $\cos(\pi n)=(-1)^n$ the given series therefore satisfies for all practical purposes the assumptions of the alternating series theorem, therefore is at least conditionally convergent.

When $0\leq x\leq{\pi\over2}$ then $\sin x\geq{2x\over\pi}$. It follows that $$\sin{\pi\over\sqrt{n}}\geq{2\over\sqrt{n}}\qquad(n\geq4)\ ,$$ and this allows to conclude that the given series is not absolutely convergent.

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