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A student recently used the series $\displaystyle\sum_{k=1}^\infty\frac{\sin^2k}{k}$ as an example of a divergent series whose terms tend to $0$. However, I'm having trouble convincing myself that this series does in fact converge. Anyone have any ideas?

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marked as duplicate by Gerry Myerson, JSchlather, Asaf Karagila, Stefan Hansen, Did Feb 14 '13 at 6:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It diverges. See this post. – David Mitra Feb 14 '13 at 5:35
The series diverges. – André Nicolas Feb 14 '13 at 5:35
Look at the terms corresponding to $k$ and $k+1$. At least one of them has the numerator bounded from below by an absolute constant. The rest should be clear (so the series, indeed, diverges). Clearly, the student is a smart guy who likes to tease his teachers a bit. So, challenge him with something tough. For instance, you can ask if the partial sums of the series $\sum_k (-1)^{[k\sqrt 2]}$ are uniformly bounded. Just don't kill him completely... – fedja Feb 14 '13 at 5:36
Whoops, didn't realize this was already asked. – Brian Fitzpatrick Feb 14 '13 at 5:41
Hmm, besides of being divergent - if I dissolve the series into a double sum (according to the power series of $\sin(x)^2/x$ )and change order of summation I get a weighted sum of zetas at negative arguments which is then nicely converging. It arrives at something like $ -0.0863018731345$ What does this tell me? – Gottfried Helms Feb 14 '13 at 5:42
up vote 6 down vote accepted

The series diverges. Notice

$$\begin{align} \sin^2(k) + \sin^2(k+1) &= \frac12(1-\cos(2k)) + \frac12(1-\cos(2k+2))\\ &= 1 - \cos(1)\cos(2k+1)\\ &\ge 1 - \cos(1)\end{align}$$

We have $$ \sum_{k=1}^{2N} \frac{\sin^2(k)}{k} = \sum_{k=1}^N\left(\frac{\sin^2(2k-1)}{2k-1} + \frac{\sin^2(2k)}{2k}\right) \ge \frac{1-\cos(1)}{2}\sum_{k=1}^N\frac{1}{k} $$ which diverges to $\infty$ as $N \to \infty$.

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@Cutiekrait: oops, fixed. – achille hui Feb 14 '13 at 5:43

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