# $\sum _{n=1}^{\infty }\sin\left(\frac{\left(-1\right)^n}{n}\right)$ Does this sum converge or diverge?

$$\sum _{n=1}^{\infty }\sin\left(\frac{\left(-1\right)^n}{n}\right)$$ Does this sum converge or diverge?

I tried this:

$$\sum _{n=1}^{\infty }\sin\left(\frac{\left(-1\right)^n}{n}\right) = \sum _{n=1}^{\infty }\sin\left(\frac{-1}{2n-1}\right) + \sum _{n=1}^{\infty }\sin\left(\frac{1}{2n}\right)$$ and I know that $$\sum _{n=1}^{\infty }\sin\left(\frac{1}{2n}\right)$$ diverges. But what about $$\sum _{n=1}^{\infty }\sin\left(\frac{-1}{2n-1}\right)$$ ?

I know that if $$\sum a_n$$ converges and $$\sum b_n$$ diverges then $$\sum \left(a_n+b_n\right)$$ diverges.

• Hint: for small angles, $\sin x$ is asymptotic to $x$. – Yves Daoust Feb 13 at 18:00
• $\sin{\frac{(-1)^n}{n}}=\frac{(-1)^n}{n}+O(n^{-3})$. – Mindlack Feb 13 at 18:01
• thank you very much! – Dr.Mathematics Feb 13 at 18:05
• Possible duplicate of how to show that $\sin(1/n)$ is a decreasing function – Lord Shark the Unknown Feb 13 at 18:05
• Note that:$$\sum_{n=1}^\infty\sin\left(\frac{(-1)^n}n\right)=\sum_{n=1}^\infty\sin\left(\frac{-1}{2n-1}\right)+\sin\left(\frac1{2n}\right)\ne\sum_{n=1}^\infty\sin\left(\frac{-1}{2n-1}\right)+\sum_{n=1}^\infty\sin\left(\frac1{2n}\right)$$ – Simply Beautiful Art Feb 14 at 12:16

Hint. Note that $$\sin\left(\frac{\left(-1\right)^n}{n}\right)=(-1)^n\sin\left(\frac{1}{n}\right)$$ and the sequence$$\{\sin(1/n)\}_{n\geq 1}$$ is positive and decreasing. Then apply Leibniz criterion.
P.S. About the odd subsequence: $$\sum _{n=1}^{\infty }\sin\left(\frac{-1}{2n-1}\right)=-\sum _{n=1}^{\infty }\sin\left(\frac{1}{2n-1}\right)=-\infty$$.