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Given the series:

$\sum_{n=1}^\infty \frac{1}{\sin n};$

does it converge? I think not but I believe that I saw somewhere in a book that it does? .... or maybe that was $\displaystyle\sum\limits_{n=1}^\infty \frac{1}{{\sin n}^2}$?

Edit: sorry, I meant $n$ .

Thanks!

Side question: Can you think of a series involving $\sin$ that supposedly gets smaller (closer to $y=0$) as it oscillates rapidly to infinity? I seem to remember reading something similar but cannot remember exactly what it was.

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    I guess that you will be also interested in [the Flint Hills Series](http://mathworld.wolfram.com/FlintHillsSeries.html) $ \sum_{n=1}^{\infty} \frac{1}{n^3 \sin^2 n} .$ It is not known whether it converges or not.2012-04-09

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I think you meant something else, ${1\over \sin x}$ is a constant...

If you meant $\sum\limits_{n=1}^\infty{1\over\sin n}$ (note you do not want to start the sum with $n=0$), you could note that $\lim\limits_{n\rightarrow\infty}{1\over\sin n}\ne 0$ (in fact, $\bigl|{1\over\sin n}\bigr|\ge 1$ for all $n$).

Recall that if $\sum\limits_{n=1}^\infty a_n$ converges, then $\lim\limits_{n\rightarrow\infty} a_n=0$.

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    @EiyrioüvonKauyf The contrapositive of the statement I wrote is: "if $\lim\limits_{n\rightarrow\infty}a_n\ne0$, then $\sum\limits_{n=1}^\infty a_n$ does not converge" (which is why some people call the statement "the Divergence Test"). So, the series diverges in this case.2012-04-08