Check if $\sum_{n=0}^\infty \frac{\sin(n)}{n}$is convergent.
I have elaborated on this question for a while and I have come up with an answer - yes, it is convergent. First, let's say that $$a_n= \frac{1}{n}$$ We can infer that $\lim a_n = 0$ and that $a_1 >a_2 > a_3> \dots$
Now, consider $$b_n = \sin(n)$$
$$\sum_{n=0}^{\infty} sin(n)$$ is clearly bounded, although I don't know how to prove it. (Any suggestions would be most welcome).
And so my reasoning has reached the seminal point - by virtue of Abel theorem - the series in question is clearly bounded. Is it correct?
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Aemilius
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How did you get that the last summation is 'clearly' bounded? – copper.hat Nov 01 '17 at 22:52
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@VidyanshuMishra, could you add a little bit more information to your comment? – Aemilius Nov 01 '17 at 22:54
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@copper.hat, I cheated a little bit and I graphed it for several thousand terms. – Aemilius Nov 01 '17 at 22:56
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That hardly constitutes 'clearly'. – copper.hat Nov 01 '17 at 22:57
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1Hint: $\sin n = \frac{1}{2i} (e^{in} - e^{-in})$. Now, calculate the partial sum using the geometric series partial sum formula. – Daniel Schepler Nov 01 '17 at 23:05