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As the title states, I'm not sure how to prove $$\sum_{n=1}^{\infty} \arctan\left(\frac{1}{F_n}\right) \arctan\left(\frac{1}{F_{n+1}}\right)=\frac{\pi^2}{8}$$ where $F_n$ representes the $n$-th fibonacci number ($F_1=1, F_2=1, F_3=2$, etc).

This question comes from an Instagram post and WolframAlpha numerically verifies the series converges to $\frac{\pi^2}{8}$ for at least $60$ decimal points. I have seen several infinite series involving arctangent and Fibonacci numbers that end up in a telescoping sum through arctangent angle addition/subtraction identities, but I'm not sure how to approach this series with the product of two arctangent functions. I'm looking for a solution that doesn't rely on knowing the series converges to $\frac{\pi^2}{8}$.

Ty.
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    Perhaps some things that will help: For $x>0$, we have $$\pi^2/8=\sum_{n=0}^\infty\frac{1}{(2n+1)^2}$$ $$\arctan(1/x)=\operatorname{arccot}(x)$$ $$\arctan(z)=\frac{z}{1+z^2}\sum_{n=0}^\infty\prod_{k=1}^n\frac{2kz^2}{(2k+1)(1+z^2)}$$ $$\arctan(z)=\sum_{n=0}^\infty \frac{2^{2n}n!^2z^{2n+1}}{(2n+1)!(1+z^2)^{n+1}}$$ $$\arctan(z)=i\sum_{n=1}^\infty\frac{1}{2n-1}\left(\frac{1}{(1+2i/z)^{2n-1}}-\frac{1}{(1-2i/z)^{2n-1}}\right)$$ – K.defaoite Apr 04 '21 at 15:29
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    This problem is mine. It is The Advanced Problem H-821 in THE FIBONACCI QUARTERLY (vol56.2 May 2018) – Hideyuki May 05 '21 at 05:55
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    For completeness I've found the link: https://www.fq.math.ca/Problems/AdvProbMay2020.pdf#page=2 – ə̷̶̸͇̘̜́̍͗̂̄︣͟ May 05 '21 at 06:37
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    @Hideyuki Sorry, the Instagram post I found it from did not cite where it was taken from. Thanks for discovering this nice problem though. – Ty. May 05 '21 at 11:35

2 Answers2

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Let $a_n=\arctan(1/F_n)$, $b_n=a_n^2$ for even $n$, and $b_n=a_{n-1}a_{n+1}$ for odd $n$ (here we assume $a_0=\pi/2$, so that $b_1=\pi^2/8$). Now I claim that $\color{blue}{a_n a_{n+1}=b_n-b_{n+1}}$. We have $$a_{n-1}-a_{n+1}=\arctan\frac{F_{n+1}-F_{n-1}}{F_{n-1}F_{n+1}+1}=\arctan\frac{F_n}{F_n^2+1+(-1)^n}$$ (with $n=1$ checked separately), hence $a_{n-1}-a_{n+1}=a_n$ for odd $n$, and \begin{align*} n\text{ is odd }&\implies b_n-b_{n+1}=a_{n-1}a_{n+1}-a_{n+1}^2=(a_{n-1}-a_{n+1})a_{n+1},\\n\text{ is even }&\implies b_n-b_{n+1}=a_n^2-a_n a_{n+2}=a_n(a_n-a_{n+2}), \end{align*} giving $a_n a_{n+1}$ in both cases. Thus $\sum_{n=1}^\infty a_n a_{n+1}=\sum_{n=1}^\infty(b_n-b_{n+1})=b_1$.

metamorphy
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COMMENT: May be this helps:

$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$

$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$

Squaring both sides of relations and subtract we get:

$$4\tan^{-1}x \tan^{-1}y=\left(\tan^{-1}\frac{x+y}{1-xy}\right)^2-\left(\tan^{-1}\frac{x-y}{1+xy}\right)^2$$

$x=\frac 1{F_n}$, and, $y=\frac 1{F_{n+1}}$

Now we use some relations between Fibonacci numbers, golden ratio $c_n$ is:

$c_n=\frac {F_{n+1}}{F_n}=\frac {F_nF_{n+1}}{F_n^2}\rightarrow F_nF_{n+1}=c_nF_n^2$

$$x+y=\frac1{F_n}+\frac1{F_{n+1}}=\frac{F_{n+2}}{c_n F_n^2}\rightarrow \frac{x+y}{1-xy}=\frac{F_nF_{n+1}F_{n+2}}{c_nF_n^2(F_nF_{n+1}-1)}$$

Similarly:

$$x-y=\frac1{F_n}-\frac1{F_{n+1}}=\frac{F_{n-1}}{c_n F_n^2}\rightarrow \frac{x-y}{1+xy}=\frac{F_{n-1}F_nF_{n+1}}{c_nF_n^2(F_nF_{n+1}+1)}$$

Now puting $F_1, F_2, F_3 \cdots$ gives us a numerical sum which must result in $\frac{\pi^2}8$.

sirous
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    may be no need to use c. In fraction we can use $ \frac {F_{n+1}}{Fn} $ instead of c.We just want to have a numerical series. – sirous Apr 08 '21 at 16:23
  • @CalvinLin, that is the point mentioned in previous comment . I said you can substitute c by $\frac{F_{N+!}}{F_N}$. The idea used in strategy is important. – sirous May 09 '21 at 04:17
  • @CalvinLin, I used c to simplify relations. You can put the ratio in it's place to continue calculations. – sirous May 09 '21 at 14:04
  • @CalvinLin, good idea, thanks – sirous May 09 '21 at 14:06