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In order to test the convergence of $\sum_{k=1}^{\infty}\frac{\sin^2k}{k^2}$, it is rather easy to comapre $ \frac{\sin^2k}{k^2}$ with $ \frac{1}{k^2}$ and use the Comparison Test ($ 0\le \frac{\sin^2k}{k^2}\le \frac{1}{k^2}$). On the other hand, since $$\lim_{k \to \infty}\frac{\frac{\sin^2k}{k^2}}{\frac{1}{k^2}} =\lim_{k \to \infty}\sin^2k $$ does not exist and Limit Coparison test cannot be used if we consider $ \frac{1}{k^2}$.

I wonder if it is possible to use Limit Comparison Test by using other $b_k$?

Edit: The Limit Comparison Test that I refer to is only consider if $\lim_{k=1}^{\infty}\frac{a_k}{b_k}$ is a positive number. See (http://mathworld.wolfram.com/LimitComparisonTest.html). May be I should rephrase my question as follow.

Does there exist $b_k$ such that $\displaystyle\lim_{k \to \infty}\frac{\left(\frac{\sin^2k}{k^2}\right)}{b_k}$ is a finite positive number and $\sum_{k=1}^{\infty}b_k$ is convergent?

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    $\begingroup$ I do not know how useful can be this but suppose you consider at the same time $ \sum_{k=1}^{\infty}\frac{\cos^2k}{k^2}$, the sum of the two sums is exactly $\frac{\pi ^2}{6}$. Just for your curiosity, the value of the sum you consider is exactly $\frac{1}{2} (\pi -1)$ $\endgroup$ Feb 15, 2014 at 4:34
  • $\begingroup$ @ClaudeLeibovici Thanks for the info. Is there any reference or hint of getting the exact sum of the series? $\endgroup$
    – pipi
    Feb 15, 2014 at 15:21
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    $\begingroup$ @pipi.May be Wolfram Alpha or another CAS. $\endgroup$ Feb 15, 2014 at 15:23
  • $\begingroup$ @ClaudeLeibovici Oh yes. I just tried Wolfram Alpha. I wonder how to calculate it in mathematical way... $\endgroup$
    – pipi
    Feb 15, 2014 at 15:33
  • $\begingroup$ Ummmmmmmh, I really don't know ! I shall work on that. Cheers. $\endgroup$ Feb 15, 2014 at 15:58

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Sure, use $b_k=\frac{1}{k^{3/2}}$. The limit is of the ratio is $0$, and since $\frac{3}{2}\gt 1$, $\sum_1^\infty \frac{1}{k^{3/2}}$ converges.

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  • $\begingroup$ Nicolas Thanks for the answer. I rephrase my question. $\endgroup$
    – pipi
    Feb 15, 2014 at 15:23
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There's a different statement of the limit comparison test that would work:

Let $\{{a_n}\}$ and $\{{b_n}\}$ be sequences of nonnegative reals. Then the following hold (the first part is what you can use):

If $\limsup_{k \to \infty}\frac{a_k}{b_k}$ is finite, then convergence of $\{{b_n}\}$ implies convergence of $\{{a_n}\}$.

If $\liminf_{k \to \infty}\frac{a_k}{b_k} > 0$, then divergence of $\{{b_n}\}$ implies divergence of $\{{a_n}\}$.)

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