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I see that quite a few questions on infinite series have been asked recently and figured why not continue the trend!

going through my lectures and textbooks, I understand $\sin^2n$ is bounded by 0 and 1, I understand that something like $\cos\pi n$ is just an alternating series in disguise, however, I cannot come up with a solid strategy for when it is just $\sin(n)$

So for example, my problem I am currently stuck on: $$\sum_{i=1}^\infty {\sin(n)\over (5+2n^2)} $$

Now I know how you would tackle something like $\sum_{i=1}^\infty \sin({1\over n}) $ but the n term isn't attached to the sin function. In my problem above I believe that is what is screwing with me.

My intuition tells me to do a comparison test to $\sum_{i=1}^\infty \sin({1\over n^2}) $

If someone could kindly explain it would be much appreciated!

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It's easier than you think. In the long run (that is for big $n$) that $\sin$ is irrelevant. –  Giuseppe Negro Nov 29 '12 at 20:52
    
Informally: $|\sin n|$ stays reasonable, $5+2n^2$ increases sort of fast, so sum is finite. –  André Nicolas Nov 29 '12 at 21:48

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Hint: The numerator is $O(1)$ and denominator is $\Omega(n^2)$, so it should converge; try comparing it with $\sum \frac{1}{n^2}$.

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that was my intuition I guess I wasn't seeing why –  treehau5 Nov 29 '12 at 20:58
    
@treehau5 Do you know about the comparison test? Observe that for any $n$ we have $\frac{\sin(n)}{5+2n^2} \leq \frac{1}{5+2n^2} \leq \frac{1}{n^2}$. –  dtldarek Nov 29 '12 at 21:01
    
yes, I guess my problem comes with the fact that the limit of sin(n) to infinity is -1 to 1 –  treehau5 Nov 29 '12 at 21:04
    
Well, $\sum_n \sin(\pi n)+\frac{1}{n^2}$ would not converge, but in your case the $\sin(n)$ is in the numerator and it becomes irrelevant as $n \to \infty$. What is important, the sequence converges absolutly. –  dtldarek Nov 29 '12 at 21:12
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Because $\frac{1}{n^2}$ converges absolutly and $\left|\frac{\sin(n)}{5+2n^2}\right| \leq \left|\frac{1}{5+2n^2}\right| \leq \left|\frac{1}{n^2}\right|$ as I pointed out before, see the aforementioned comparison test for more details. –  dtldarek Nov 29 '12 at 21:25

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