TITLE: Infinite Series $\sum\limits_{n=0}^{\infty}\arctan(\frac{1}{F_{2n+1}})$
QUESTION [7 upvotes]: How can I find the value of the following sum?
$$\sum_{n=0}^{\infty}\arctan(\frac{1}{F_{2n+1}})$$
$F_n$ is the Fibonacci number.($F_1=F_2=1$)

REPLY [9 votes]: We can prove by induction that
$$
\sum_{n=0}^k\arctan\frac1{F_{2n+1}}=\arctan F_{2k+2}.
$$
The base case $k=0$ is immediate, because $F_1=F_2=1$.
OTOH by induction hypothesis
$$
\sum_{n=0}^{k+1}\arctan\frac1{F_{2n+1}}=\arctan F_{2k+2}+\arctan\frac1{F_{2k+3}}.
$$
It follows from the formula for the tangent of the sum of two angles (careful about the overflow!)
$$
\arctan x+\arctan y\equiv \arctan\frac{x+y}{1-xy}\pmod\pi.
$$
Here $x=F_{2k+2}$, $y=1/F_{2k+3}$, and thus
$$
\begin{aligned}
\frac{x+y}{1-xy}&=\frac{F_{2k+2}F_{2k+3}+1}{F_{2k+3}-F_{2k+2}}\\
&=\frac{F_{2k+2}F_{2k+3}+1}{F_{2k+1}}=F_{2k+4}
\end{aligned}
$$
by the identity $F_{2k+4}F_{2k+1}=1+F_{2k+2}F_{2k+3}$ completing our induction step (see the link given by Dan Shved or prove this identity by induction).
As $F_n\to\infty$ it follows that the limit is $\pi/2$.