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Edit: I am seeking a solution that uses only calculus and real analysis methods -- not complex analysis. This is an old advanced calculus exam question, and I think we are not allowed to use any complex analysis that could make the problem statement a triviality.

Show that the series

$$\sum_{n=2}^{\infty} \frac{\sin(n)}{\log(n)}$$

converges.

Any hints or suggestions are welcome.

Some thoughts:

The integral test is not applicable here, since the summands are not positive.

The Dirichlet test does seem applicable either, since if I let 1/log(n) be the decreasing sequence, then the series of sin(n) does not have bounded partial sums for every interval.

Thanks,

User001
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    How do you know that it converges? It doesn't seem like it would converge in my mind. It definitely doesn't converge absolutely. – cheesyfluff Nov 24 '15 at 04:21
  • Hi @cheesyfluff - the problem statement asks to show that it converges. But you're right in some sense, I have worked on problems that have asked, "find all solutions to...", and the answer was indeed that there were no solutions ... – User001 Nov 24 '15 at 04:23
  • You have 3-4 positive terms followed by 3-4 negative terms and the number of positive and negative terms averages out. So you sequence is like an alternating conditionally convergent series - converges conditionallu. – A.S. Nov 24 '15 at 04:46

1 Answers1

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Note that

$$\left|\sum_{k=1}^n \sin k\right|= \frac{|\sin(n/2)\sin[(n+1)/2]|}{\sin(1/2)}\leqslant \frac{1}{\sin(1/2)}.$$

Derivation:

$$\begin{align} 2 \sin(1/2)\sum_{k=1}^n\sin k &= \sum_{k=1}^n2 \sin(1/2)\sin k \\ &= \sum_{k=1}^n2 \sin(1/2)\cos (k +\pi/2) \\ &= \sum_{k=1}^n[\sin(k + 1/2 + \pi/2)-\sin (k -1/2 + \pi/2)] \\ &= \sin(n + 1/2 + \pi/2) - \sin(1/2 + \pi/2) \\ &= 2 \sin(n/2)\cos[(n+1)/2 +\pi/2] \\ &= 2 \sin(n/2)\sin[(n+1)/2] \end{align} \\ \implies \sum_{k=1}^n\sin k = \frac{\sin(n/2)\sin[(n+1)/2]}{\sin(1/2)}$$

RRL
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  • How did you arrive at that first equality? – ASKASK Nov 24 '15 at 04:18
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    @ASKASK One can derive that by noting that $\sin(k)$ is the sum of two geometric series: $$\sin(k)=\frac{1}2\left[(e^i)^k-(e^{-i})^k\right]$$ and then using standard formulas for sum over geometric series. – Milo Brandt Nov 24 '15 at 04:19
  • Hi @MiloBrandt, now that you provided this suggestion, along with RRL's derivation, I am concerned about whether I can use it - and I will edit my question to address this concern. The question is from an old advanced calculus exam -- and so I believe that (on exam day) we are not allowed to utilize complex analysis methods to solve this problem. A similar problem was to show the series of cos(n)/log(n) converges (I was not able to solve this one either), and again I think that one must stick with calculus / real analysis methods only ... – User001 Nov 24 '15 at 04:32
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    @Lebron James: The Dirichlet test is applicable for convergence of the series You can also derive that equality without complex variables using trig identities and a telescoping sum. – RRL Nov 24 '15 at 04:37
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    @LebronJames One could prove the identity mentioned here by induction and trigonometric identities. I wouldn't recommend it, but you can. You can also prove it by considering the sum of a bunch of unit vectors on the plane separated by a constant angle (for this, consider the sum minus the sum rotated by this angle). Also, I would suspect that if you're asked to avoid complex analysis, that's more of a "Don't use contour integrals to solve this" than a "Don't think about complex numbers at all" – Milo Brandt Nov 24 '15 at 04:39
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    @LebronJames Good to see you're still here during the NBA season. To expand on the last comment by RRL, simply note $$\sin (1)\sin (n)=\frac12\left(\cos (n-1)-\cos (n+1)\right)$$Then, telescope. – Mark Viola Nov 24 '15 at 05:10
  • Hi @MiloBrandt, thanks so much for your comments -- yes, I am struggling a bit with going back to prove it using trig identities. I have gotten better with going back to trig identities recently, as it seems to show up a lot on exams, but this one is particularly tough to spot -- and prove. – User001 Nov 24 '15 at 23:11
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    Thanks so much for your answer and derivation @RRL :-) – User001 Nov 24 '15 at 23:12
  • Hi @Dr.MV! Nice to hear from you :-) I see you've been quite busy contributing to MSE - thanks so much for your work. When I am stuck on a problem, I often find a great write-up by you...somewhere on this site for a related question. Yep, with Kyrie Irving and Kevin Love back from injuries, I am able to allocate some more time to mathematics ;) – User001 Nov 24 '15 at 23:15
  • Hi @Dr.MV, I was wondering whether you possibly knew of a "more natural" way of solving this problem, instead of knowing the key step above of multiplying by a sin(1/2), and then going for the product-to-sum trig formulas. E.g., would there be a way to solve it using Taylor series / Taylor's remainder theorem / Taylor expansion / and asymptotics (big-O) arguments? – User001 Nov 24 '15 at 23:19
  • In particular, I once saw something like this for the series of cosn/logn, and the technique was something like looking at the difference between the sum and an integral. I don't know what the source was, so I can't find it anymore. What do you think? Thanks so much, @Dr.MV, – User001 Nov 24 '15 at 23:19
  • Hi @MiloBrandt - so today I saw several similar questions like this on MSE, where the users also asked for derivations using only real-valued functions. This concerned me just a bit, and I reached out to a professor in our dept and am just waiting to hear back from him. But I just finished working out the (simple) details to show the boundedness from your suggestion of using the complex sine formula. It's nice and very natural to think about it in this way. Thanks so much, – User001 Nov 25 '15 at 03:24
  • @RRL, in your answer there is a typo indeed $\sin k=\mathbf{\color{red}-}\cos(k+\pi/2)$. – Angelo Sep 22 '25 at 03:56