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While trying to evaluate for $n \ge 2$ integer the sum $$\sum_{r=0}^{n-2} 2^r \tan \frac{\pi}{2^{n-r}}$$ I remembered that, for any $k\in\mathbb{N}$, the following identity holds $$\sin x=2^k \sin \frac{x}{2^k} \prod_{j=1}^k \cos \frac{x}{2^j}$$ So, for $k=n-r$, with $n\ge 2$, I get that $$2^{n-r} \sin \frac{x}{2^{n-r}}=\sin x \left(\prod_{j=1}^{n-r} \cos \frac{x}{2^j}\right)^{-1}\iff 2^r \sin \frac{x}{2^{n-r}}=2^{2r-n}\sin x \left(\prod_{j=1}^{n-r} \cos \frac{x}{2^j}\right)^{-1}$$ Substitung $x=\pi$, the right hand side is $0$ and so I get that $$0=2^r \sin \frac{\pi}{2^{n-r}}$$ And this implies that $2^r \tan \frac{\pi}{2^{n-r}}=0$ and so the sum is $0$.

However this is surely wrong, for two reasons: because it is not true that it is always $0=2^r \sin \frac{\pi}{2^{n-r}}$ (it is enough to substitute $r=0$ and $n=2$) and, if I'm not wrong, for any $0 \le r \le n-2$ and for any $n \ge 2$ it is $0<\frac{\pi}{2^{n-r}}<\frac{\pi}{2}$ and so for any $0 \le r \le n-2$ and for any $n \ge 2$ it is $2^r \tan \frac{\pi}{2^{n-r}}>0$, hence the sum is surely positive being a sum of positive numbers.

Indeed, using the identity $\tan \theta=\cot \theta -2\cot 2\theta$, is it possible to prove that for any $n \ge 2$ integer $$\sum_{r=0}^{n-2} 2^r \tan \frac{\pi}{2^{n-r}}=\cot \frac{\pi}{2^n}$$

I suspect that the problem is when I divide both sides for $\prod_{j=1}^{n-r} \cos \frac{x}{2^j}$, because it can be $0$ and so the identity doesn't hold anymore: $$\prod_{j=1}^{n-r} \cos \frac{x}{2^j}=0 \iff \cos \frac{x}{2^j}=0, \ j\in\{1,...,n-r\} \iff x=2^{j-1} \pi+2^j h\pi, \ j\in\{1,...,n-r\}, h\in\mathbb{Z}$$ And, for example, for $j=1$ and $h=0$ the product is indeed $0$ and so I've divided by $0$ at least one time. Is this the reason why my reasoning doesn't work?

Moreover, I would like to ask what really is $$\lim_{n \to \infty} \sum_{r=0}^{n-2} 2^r \tan \frac{\pi}{2^{n-r}}$$ Can this be considered a series? I'm unsure because usually in a series the index that goes to infinity is not both in the upper summation bound and in the sequence we are summing, like in this case. So I don't know if I can use any criteria I know to study it, instead of just hoping that, like in this situation, I can evaluate explicitly the sum and say that $$\sum_{r=0}^{n-2} 2^r \tan \frac{\pi}{2^{n-r}}=\lim_{n\to\infty} \cot \frac{\pi}{2^n}=\infty$$

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    $\begingroup$ Indeed, when $x = \pi$ then both $\sin \pi$ and $\cos\frac{\pi}{2}$ are zero. And as you correctly computed, your limit is $\infty$. Also, the sum itself is not a partial sum of an infinite series because the summands depend on the variable $n$. $\endgroup$ Commented Nov 21, 2021 at 16:25
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    $\begingroup$ Let's look at each line when $x = \pi$. Your original equation is $0=0$ when $x = \pi$. You then divide both sides by zero when you divide by the cosine product. Unsurprisingly, illogic follows. Alternatively, as soon as you divide by the cosine product, you acquire the condition $x \not\in \{k\pi \mid k \in \Bbb{Z}\}$. $\endgroup$ Commented Nov 21, 2021 at 16:25
  • $\begingroup$ Thanks to both Sangchul Lee and Eric Towers for your comments, now everything is clear. $\endgroup$
    – Gwyn
    Commented Nov 21, 2021 at 23:24

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