# Prove $\sum_{n=1}^{\infty} \arctan\left(\frac{1}{F_n}\right) \arctan\left(\frac{1}{F_{n+1}}\right)=\frac{\pi^2}{8}$

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 $$\frac{\pi^2}{8}$$.

• Perhaps some things that will help: For $x>0$, we have $$\pi^2/8=\sum_{n=0}^\infty\frac{1}{(2n+1)^2}$$ $$\arctan(1/x)=\operatorname{arccot}(x)$$ $$\arctan(z)=\frac{z}{1+z^2}\sum_{n=0}^\infty\prod_{k=1}^n\frac{2kz^2}{(2k+1)(1+z^2)}$$ $$\arctan(z)=\sum_{n=0}^\infty \frac{2^{2n}n!^2z^{2n+1}}{(2n+1)!(1+z^2)^{n+1}}$$ $$\arctan(z)=i\sum_{n=1}^\infty\frac{1}{2n-1}\left(\frac{1}{(1+2i/z)^{2n-1}}-\frac{1}{(1-2i/z)^{2n-1}}\right)$$ Commented Apr 4, 2021 at 15:29
• This problem is mine. It is The Advanced Problem H-821 in THE FIBONACCI QUARTERLY (vol56.2 May 2018) Commented May 5, 2021 at 5:55
• For completeness I've found the link: fq.math.ca/Problems/AdvProbMay2020.pdf#page=2 Commented May 5, 2021 at 6:37
• @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.
Commented May 5, 2021 at 11:35

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 $$\sum_{n=1}^\infty a_n a_{n+1}=\sum_{n=1}^\infty(b_n-b_{n+1})=b_1$$.

• Cool answer, but this is missing a LOT of steps. How do we use the multiplication formula for Fibonacci numbers when they are wrapped up in an arctangent? Commented Apr 4, 2021 at 16:40
• @K.defaoite: Wait a moment ;) Commented Apr 4, 2021 at 16:42
• Thanks. I'm actually looking for a solution that arrives at $\frac{\pi^2}{8}$ without knowing this is the value that the series converges to. The wording of my question was probably poor.
– Ty.
Commented Apr 4, 2021 at 16:51
• @Ty.: I've rewritten it in a "telescopic" spirit. Perhaps it looks better this way. Commented Apr 4, 2021 at 17:14
• @ZackNi: See this. Commented May 5, 2021 at 6:35

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 $$\frac{\pi^2}8$$.

• 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. Commented Apr 8, 2021 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. Commented May 9, 2021 at 4:17
• @CalvinLin, I used c to simplify relations. You can put the ratio in it's place to continue calculations. Commented May 9, 2021 at 14:04
• @CalvinLin, good idea, thanks Commented May 9, 2021 at 14:06