I want to know about the convergence or divergence of the following series:

$$\sum \sin (a_n) $$



The tests that I tried were inconclusive. Is it possible to know?

Or, more generally, does this series converge for all conditionally convergent series given by the sequence $a_n$? I know this result holds for any absolutely convergent series (here), but I want to know if it works for any convergent series or if there's an explicit counterexample taking conditionally convergent series.

EDIT: as pointed out at the answer section, my example converges, but the general case is still unanswered.


For the first part

Does $\sum\limits_{n=1}^\infty \sin\frac{(-1)^n}{n}$ converges?

The answer is YES by Alternating series test.

For the second part

Does $\sum\limits_{n=1}^\infty \sin a_n$ converges for all conditional converging series $\sum\limits_{n=1}^\infty a_n$ ?

The answer is NO. Consider the series

$$a_n = \frac{\epsilon_n}{\lceil n/3 \rceil ^{1/3}} \quad\text{ where }\quad \epsilon_n = \begin{cases} +1, & n \not\equiv 0 \pmod 3\\ -2, & n \equiv 0 \pmod 3 \end{cases}$$

If we group the terms in units of three, one find

$$a_{3k-2} + a_{3k-1} + a_{3k} = 0$$ and the series $\sum\limits_{n=1}^\infty a_n$ converges conditionally. However, $$\sin a_{3k-2} + \sin a_{3k-1} + \sin a_{3k} = 2\sin\frac{1}{\sqrt[3]{k}} - \sin\frac{2}{\sqrt[3]{k}} = \frac{1}{k} + O\left(\frac{1}{k^{5/3}}\right)$$

This implies the partial sums $\sum\limits_{n=1}^N a_{n}$ behaves roughly as $\sum\limits_{k=1}^{\lfloor N/3 \rfloor} \frac{1}{k} \sim \log\frac{N}{3}$ for large $N$ and hence the corresponding series diverges.


$\sin(a_n)=(-1)^n\sin(\frac{1}{n})$ so you are dealing with an alternating series. Then the fact that $\sin(\frac{1}{n})$ is monotonically decreasing and converges to $0$ justifies the conclusion that the series is converging. The so-called Leibniz-test is applied.


For $x$ close to $0$, we have $\frac{|x|}{2}\leq |\sin{x}|\leq |x|$. Does that give you any ideas?

Edit: can be done way simpler, just noticed. $\sin(-x)=-\sin(x)$, and the sequence $\frac{1}{n}$ has the limit $0$. Now, if I were to mention a guy called Leibniz, we are almost done.


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