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My question is as follows.

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?

We suppose that we do not know that $\displaystyle\sum_{k =1}^{\infty}{\left(\frac{\sin^2k}{k^2}\right)}$ is convergent.

This question is not yet answered so far but there is a related question here.

Community
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pipi
  • 2,461
  • I think its safe to say that we can assume we know $\lim_{k\to \infty} \frac{sin^2 k}{k^2}$ is convergent since clearly it is a necessary condition that the elements of a sequence tend to zero for its series to converge. And we know in fact that $\sum_{k\in \mathbb{N}} \frac{sin^2 k}{k^2} < \sum_{k\in \mathbb{N}} \frac{1}{k^2} = \frac{\pi}{6}$, and hence is convergent.... – Squirtle Feb 17 '14 at 05:28
  • yeah.... tired.... confused it with 2^n. Thanks Also, I suppose its safe to say we should know this naturally. – Squirtle Feb 17 '14 at 05:31

1 Answers1

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Simply take $b_k=\sin^2(k)/k^2$.

Clayton
  • 24,751
  • thanks for the answer. How if we suppose that we do not know that $\sum_{k =1}^{\infty}{\left(\frac{\sin^2k}{k^2}\right)}$ is convergent. – pipi Feb 17 '14 at 05:26
  • Agreed.... I don't think this is obvious, perhaps there is a more "natural/elementary" answer. – Squirtle Feb 17 '14 at 05:33