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$$ \sum _{n=1}^{\infty }\sin\left(\frac{\left(-1\right)^n}{n}\right) $$ 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.

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  • $\begingroup$ Hint: for small angles, $\sin x$ is asymptotic to $x$. $\endgroup$
    – user65203
    Feb 13, 2019 at 18:00
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    $\begingroup$ $\sin{\frac{(-1)^n}{n}}=\frac{(-1)^n}{n}+O(n^{-3})$. $\endgroup$
    – Aphelli
    Feb 13, 2019 at 18:01
  • $\begingroup$ thank you very much! $\endgroup$
    – Dr.Mathematics
    Feb 13, 2019 at 18:05
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    $\begingroup$ Possible duplicate of how to show that $\sin(1/n)$ is a decreasing function $\endgroup$
    – Angina Seng
    Feb 13, 2019 at 18:05
  • $\begingroup$ Note that:$$\sum_{n=1}^\infty\sin\left(\frac{(-1)^n}n\right)=\sum_{n=1}^\infty\sin\left(\frac{-1}{2n-1}\right)+\sin\left(\frac1{2n}\right)\ne\sum_{n=1}^\infty\sin\left(\frac{-1}{2n-1}\right)+\sum_{n=1}^\infty\sin\left(\frac1{2n}\right)$$ $\endgroup$
    – Simply Beautiful Art
    Feb 14, 2019 at 12:16

1 Answer 1

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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: $\sum _{n=1}^{\infty }\sin\left(\frac{-1}{2n-1}\right)=-\sum _{n=1}^{\infty }\sin\left(\frac{1}{2n-1}\right)=-\infty$.

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  • $\begingroup$ thank you very much! $\endgroup$
    – Dr.Mathematics
    Feb 13, 2019 at 18:05

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