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Context: I evaluated some sums related to $1/\pi^{2}$ that for a reason I don't know I can't find in the literature.

$$(1)\hspace{.5cm}{1\over 2\sqrt{2}}+{1\over 2^{3}\sqrt{2+\sqrt{2}}}+{1\over 2^{5}\sqrt{2+\sqrt{2+\sqrt{2}}}}+\cdots=1-{8\over \pi^{2}},$$

$$(2)\hspace{.5cm}{1/4\over 2+\sqrt{3}}+{1/16\over 2+\sqrt{2+\sqrt{3}}}+{1/64\over 2+\sqrt{2+\sqrt{2+\sqrt{3}}}}+\cdots=1-{9\over \pi^{2}},$$ $$(3)\hspace{.5cm}{1/4\over 2+\sqrt{2+\varphi}}+{1/16\over 2+\sqrt{2+\sqrt{2+\varphi}}}+{1/64\over 2+\sqrt{2+\sqrt{2+\sqrt{2+\varphi}}}}+\cdots={1+\varphi}-{25\over \pi^{2}},$$ where $\varphi=\frac{1+\sqrt{5}}{2}.$

Note that this series are related to Viète's formulas.
Question: Do you know any reference where these formulas are shown? Can you prove them with a general result?

Anne Bauval
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    How did you "evaluate" these sums? – Anne Bauval Feb 05 '23 at 11:44
  • @AnneBauval I'm still waiting for an answer to some references to this sums if there exists. After this I will post the answer with full details. –  Feb 05 '23 at 11:50
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    All series should be telescoping sum. For $(2)$ and $(3)$, they can be solved using identity $\frac{1}{\sin^2(\theta)} - \frac{1}{4\sin^2\frac{\theta}{2}} = \frac{1}{4\cos^2\frac{\theta}{2}} = \frac{1}{2+2\cos(\theta)}$. Havn't figure out the correct decomposition for $(1)$... – achille hui Feb 05 '23 at 14:19
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    @achille hui Excellent remark ! Do you intend to give a full answer ? – Jean Marie Feb 06 '23 at 10:01
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    I don't see how this series can be related to Viète's formulas. Please, write down the "answer" you intended to give ; even if it isn't complete, it will be fruitful for the community, just as the very interesting track given by achille hui above. – Jean Marie Feb 06 '23 at 10:15
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    @JeanMarie probably not, still stuck on $(1)$... – achille hui Feb 06 '23 at 12:41
  • @achillehui you are right, the three sums are telescopic, but they can be deduced from a general formula (the hint: Viète's formulas) should be useful. –  Feb 06 '23 at 13:29
  • Please stop modifying your post after it has been answered. – Anne Bauval Mar 05 '23 at 07:55

1 Answers1

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I will consider here only the third series with $\varphi$ the golden ratio.

I will establish it as an application of the following identity, justifying the presence of a $\frac{1}{\pi^2}$ term :

$$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$

Its proof by different means is given in the answers to this question. Among them, one finds a proof using Viète's formulas...

Indeed, taking $x=\frac{\pi}{5}$ in (1), one gets, in the RHS :

$$\frac{1}{\sin^2 \frac{\pi}{5}}-\frac{25}{\pi^2}=\frac{4}{3-\pi}-\frac{25}{\pi^2}$$

On the LHS, detailing the first terms :

$$\color{red}{\underbrace{\frac{1}{4\cos^2\frac{\pi}{5 \times 2}}}_{= \ \frac{1}{2+\varphi}}}+\frac{1}{4^2\cos^2\frac{\pi}{5 \times 2^2}}+\frac{1}{4^3\cos^2\frac{\pi}{5 \times 2^3}}+\cdots$$

Why is the first term highlighted ? Because it is the single term which is not present in the series you have given. All the other terms are exactly the right ones :

Indeed, one can verify that :

$$4 \cos^2 \left(\frac{\pi}{5.2^k}\right)=\begin{cases}2+\varphi&(k=1)\\ 2+\sqrt{2+\varphi}&(k=2)\\2 + \sqrt{2+\sqrt{2+\varphi}}&(k=3)\\ \cdots \end{cases}$$

due to relationship :

$$\cos\left(\frac{\pi}{5.2^{k+1}}\right)=\frac12\left(1+\cos\left(\frac{\pi}{5.2^{k}}\right)\right) \iff $$

$$4\left(\cos\left(\frac{\pi}{5.2^{k+1}}\right)\right)^2=2+2 \cos\left(\frac{\pi}{5.2^{k}}\right) $$

It remains for completing the proof to subtract to the RHS the "added term" :

$$\frac{4}{3-\pi}-\frac{25}{\pi^2}-\color{red}{\frac{1}{2+\varphi}}=1+\varphi-\frac{25}{\pi^2}$$

as desired (using relationshup $\varphi^2=\varphi+1$, etc.).

Remark : the different special values such as $\sin \frac{\pi}{5}$, etc. can be found for example here.

Jean Marie
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