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According to WolframAlpha this series converges, but I can't find out how to properly use the limit comparison test with it. Can anyone at least tell me what my $b_n$ might be? $$\sum_{n=1}^\infty\sqrt{n}\cdot\sin^2\left(\frac{\pi}n\right)$$

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I edited your post as I understood it. If there are any mistakes - feel free to edit it. Please use $\LaTeX$ to typeset math. You can find a nice guide here: meta.math.stackexchange.com/questions/5020/… –  Dennis Gulko Mar 21 '13 at 8:38
    
Thank you! Hopefully someone will follow up with an answer. –  Paul Mar 21 '13 at 8:39
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2 Answers 2

Remember that $\lim_{x\to0}\frac{\sin x}x=1$, hence when you see $\sin^2\left(\frac{\pi}n\right)$,
you should immediately think of $\left(\frac{\pi}n\right)^2$. So try $b_n=\sqrt{n}\left(\frac{1}n\right)^2=\frac1{n\sqrt n}$.
(and use the limit comparison test)

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Great! Thanks a bunch. –  Paul Mar 21 '13 at 8:56
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For all $x$, we have $$ \sin^2(x)\le x^2 $$ Thus, $$ \left|\sum_{n=1}^\infty\sqrt{n}\cdot\sin^2(\pi/n)\right|\le\sum_{n=1}^\infty\frac{\pi^2}{n^{3/2}} $$ which converges by the $p$-test ( $\frac32\gt1$ ).

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