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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 $n$ .


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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$1/\sin x$ is a constant... If you meant $1/\sin n$, you could show that $\lim\limits_{n\rightarrow\infty}(1/\sin n)\ne 0$. – David Mitra Apr 8 '12 at 20:48
Are you sure this is what you want to ask? The general term of the series does not depend on $n$. – Julián Aguirre Apr 8 '12 at 20:49
For a real number $t$ such that $\sin t \neq 0$ we have $\frac 1{\sin t}\in \mathbb R_{\leq -1}\cup\mathbb R_{\geq 1}$, so theses series cannot converge. – Davide Giraudo Apr 8 '12 at 20:53
Maybe you mean series like $\sum_{n\geq 1}\frac 1{n^2\sin n}$. – Davide Giraudo Apr 8 '12 at 20:58
That would basically be a $1/n$ convergence though, no? – Eiyrioü von Kauyf Apr 8 '12 at 20:59
up vote 9 down vote accepted

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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Hmm... Actually I thought if \lim_{n \to \infty} n = 0 then all you know is that it is not divergent not if it is convergent or not. So can you elaborate on your third line? With regard to $\frac{1}{sin n}$ never converging to 0 I can agree with that. – Eiyrioü von Kauyf Apr 8 '12 at 20:58
@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. – David Mitra Apr 8 '12 at 21:06

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