This appeared on the exercises sheet for a "Numerical Series" chapter of a university course: "Determine the nature and the possible sum of the numerical series". Among 18 examples I solved (or found solution for) 17 and was defeated by this: $$\sum_{n=0}^{+\infty}\ln\left(2\cos\frac{\alpha}{2^n}-1\right)$$ The question as it appears doesn't even specify anything about $\alpha$, so I assume "Generality of Algebra" here, but I can't even start or see a single trigonometric identity that produces a telescoping series.
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- 9$\begingroup$ The identities $$2\cos x-1=\frac{\cos(3x/2)}{\cos(x/2)}\qquad\text{and}\qquad\frac{\sin x}{x}=\prod_{n=1}^{\infty}\cos\left(\frac{x}{2^n}\right)$$ would help. Let me know if you need a full-fledged answer! $\endgroup$Sangchul Lee– Sangchul Lee2025-11-20 14:21:24 +00:00Commented Nov 20 at 14:21
- $\begingroup$ @SangchulLee Thank you so much! This is already a full-fledged answer. I will write the answer and credit you. The second formula you provided is the next question in my exercises sheet (which I already found solution for). It is not homework and I solved everything in the sheet except the current question, but my instructor refused to share the solution when I requested anyway, curiosity would've driven me and my colleagues crazy tbh. $\endgroup$zaknenou– zaknenou2025-11-20 15:09:10 +00:00Commented Nov 20 at 15:09
- $\begingroup$ @SangchulLee sorry to bother you again, but I've spent some time thinking about the solution, but I can't see how did the transformation occur to you, it doesn't seem like one of the standard identities at all. I mean $2\cos x-1=\frac{\cos(3x/2)}{\cos(x/2)}$, is there some hint for that ? I'm surprised you were so quick with the solution. $\endgroup$zaknenou– zaknenou2025-11-22 15:30:00 +00:00Commented Nov 22 at 15:30
- 1$\begingroup$ The identity is best explained in the complex domain. Indeed, $$\begin{align*}2\cos x-1&=e^{ix}-1+e^{-ix}\\&=e^{-ix}(1 - e^{ix} + e^{2ix})\\ &= e^{-ix} \frac{1 + e^{3ix}}{1 + e^{ix}}\\ &= \frac{\cos(3x/2)}{\cos(x/2)}. \end{align*}$$ I originally tried to write everything in terms of the complex exponential and then see what will happen, and I was indeed lucky that this approach worked. :) $\endgroup$Sangchul Lee– Sangchul Lee2025-11-22 18:55:34 +00:00Commented Nov 22 at 18:55
- $\begingroup$ @SangchulLee Thanks, this fills the gap. $\endgroup$zaknenou– zaknenou2025-11-23 13:59:01 +00:00Commented Nov 23 at 13:59
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2 Answers
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4 we have : \begin{align*} A &= \sum_{n=0}^{\infty}\ln\left(2\cos\frac{\alpha}{2^n}-1\right) \\&= \ln \prod_{n=0}^{\infty} \left({2\cos\frac{\alpha}{2^n}-1}\right) \\&= \ln \prod_{n=0}^{\infty} \frac{2\cos\left({\frac{\alpha}{2^{n-1}}}\right)+1}{2\cos\left({\frac{\alpha}{2^{n}}}\right)+1} \\&= \ln(2\cos(2\alpha)+1) \end{align*}
- 1$\begingroup$ I'm sorry if this is a stupid question, but I can't find how the 3rd line is deduced ? $\endgroup$zaknenou– zaknenou2025-11-20 15:38:15 +00:00Commented Nov 20 at 15:38
- $\begingroup$ I guess you forgot the factor $3$, since $$\prod_{n=0}^{N}\frac{2\cos(\alpha 2^{1-n})+1}{2\cos(\alpha 2^{-n})+1}=\frac{2\cos(2\alpha)+1}{2\cos(\alpha 2^{-N})+1}$$ and the denominator converges to $3$. Other than that, a very nice observation indeed! $\endgroup$Sangchul Lee– Sangchul Lee2025-11-20 16:34:31 +00:00Commented Nov 20 at 16:34
- $\begingroup$ @BoweiTang Even that I couldn't prove, Are you sure the identity is true ?: ibb.co/hk6XF47 $\endgroup$zaknenou– zaknenou2025-11-20 20:16:42 +00:00Commented Nov 20 at 20:16
- $\begingroup$ @zaknenou Oh, sorry, I didn't verify it carefully. It seems that this answer is not correct. $\endgroup$Bowei Tang– Bowei Tang2025-11-21 02:12:02 +00:00Commented Nov 21 at 2:12
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Following user Sangchul Lee (I am only following his hints from the comment here) we'll be using the identity:
$$2\cos x-1=\frac{\cos(3\frac{x}{2})}{\cos(\frac{x}{2})}$$ so let's prove it quick:
\begin{align*} \cos 3x &=\cos 2x \cos x - \sin 2x \sin x \\&= \cos 2x \cos x - 2\sin^2 x \cos x \\&= \left(\cos 2x- 2\frac{1-\cos 2x}{2} \right) \cos x \\&= \left(2\cos 2x- 1 \right) \cos x \end{align*} thus: $2\cos 2x-1=\frac{\cos 3x}{\cos x}$ and by substituting $\frac{x}{2}$ we get the identity. In our case it becomes: $$2\cos \left(\frac{\alpha}{2^{n}}\right)-1=2\cos \left(2\times \frac{\alpha}{2^{n+1}}\right)-1=\frac{\cos \left(3 \times \frac{\alpha}{2^{n+1}}\right)}{\cos \frac{\alpha}{2^{n+1}}}$$ Now to the main sum: \begin{align*} \sum_{n=0}^{+\infty}\ln\left(2\cos\frac{\alpha}{2^n}-1\right)&=\ln\left(\prod_{n=0}^{+\infty}\left(2\cos\frac{\alpha}{2^n}-1\right)\right) \\&= \ln\left(\prod_{n=0}^{+\infty}\frac{\cos \left(3 \times \frac{\alpha}{2^{n+1}}\right)}{\cos \frac{\alpha}{2^{n+1}}}\right) \\&= \ln\left(\frac{\prod_{n=0}^{+\infty} \cos \left( \frac{3\alpha}{2^{n+1}}\right)}{\prod_{n=0}^{+\infty}\cos \frac{\alpha}{2^{n+1}}}\right) \\&= \ln\left(\frac{\prod_{n=1}^{+\infty} \cos \left( \frac{3\alpha}{2^{n}}\right)}{\prod_{n=1}^{+\infty}\cos \frac{\alpha}{2^{n}}}\right) \\&= \ln\left(\frac{\frac{\sin \left( 3\alpha \right)}{3\alpha}}{\frac{\sin \left( \alpha \right)}{\alpha}}\right) \\&= \ln\left(\frac{\sin \left( 3\alpha \right)}{3 \sin \left( \alpha \right)}\right) \end{align*} The fifth line follows from this identity (here is a specific thread) : $$\prod_{n=1}^{\infty}\cos\left(\frac{x}{2^n}\right)=\frac{\sin x}{x}$$
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