\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We aim to show that for all \(n \geq 0\), \[ F_n F_{n + 3} - F_{n + 1} F_{n + 2} = F_{n + 2} F_{n + 5} - F_{n + 3} F_{n + 4}. \]
Notice that \begin {align*} \RHS & = F_{n + 2} F_{n + 5} - F_{n + 3} F_{n + 4} \\ & = F_{n + 2} (F_{n + 3} + F_{n + 4}) - F_{n + 3} (F_{n + 2} + F_{n + 3}) \\ & = F_{n + 2} F_{n + 4} - F_{n + 3} F_{n + 3} \\ & = F_{n + 2} (F_{n + 2} + F_{n + 3}) - F_{n + 3} (F_{n + 1} + F_{n + 2}) \\ & = F_{n + 2} F_{n + 2} - F_{n + 3} F_{n + 1} \\ & = F_{n + 2} (F_{n + 3} - F_{n + 1}) - F_{n + 3} (F_{n + 2} - F_{n}) \\ & = F_{n} F_{n + 3} - F_{n + 1} F_{n + 2} \\ & = \LHS \end {align*}
and set \(n = 0\) shows exactly what is desired.
By the lemma in the previous part, the problem reduces to two cases are when \(n\) is odd and when \(n\) is even.
Using the tangent formula for sum of angles, we have \begin {align*} \arctan \left (\frac {1}{F_{2r + 1}}\right ) + \arctan \left (\frac {1}{F_{2r + 2}}\right ) & = \arctan \left (\frac {\frac {1}{F_{2r + 1}} + \frac {1}{F_{2r + 2}}}{1 - \frac {1}{F_{2r + 1}} \cdot \frac {1}{F_{2r + 2}}}\right ) \\ & = \arctan \left (\frac {F_{2r + 1} + F_{2r + 2}}{F_{2r + 1} F_{2r + 2} - 1}\right ) \\ & = \arctan \left (\frac {F_{2r + 3}}{F_{2r + 1} F_{2r + 2} + (F_{2r} F_{2r + 3} - F_{2r + 1} F_{2r + 2})}\right ) \\ & = \arctan \left (\frac {F_{2r + 3}}{F_{2r} F_{2r + 3}}\right ) \\ & = \arctan \left (\frac {1}{F_{2r}}\right ), \end {align*}
as desired.
Hence, we have \[ \arctan \left (\frac {1}{F_{2r + 1}}\right ) = \arctan \left (\frac {1}{F_{2r}}\right ) - \arctan \left (\frac {1}{F_{2r + 2}}\right ), \] and therefore \begin {align*} \sum _{r = 1}^{\infty } \arctan \left (\frac {1}{F_{2r + 1}}\right ) & = \sum _{r = 1}^{\infty } \arctan \left (\frac {1}{F_{2r}}\right ) - \sum _{r = 1}^{\infty } \arctan \left (\frac {1}{F_{2r + 2}}\right ) \\ & = \sum _{r = 1}^{\infty } \arctan \left (\frac {1}{F_{2r}}\right ) - \sum _{r = 2}^{\infty } \arctan \left (\frac {1}{F_{2r}}\right ) \\ & = \arctan \left (\frac {1}{F_{2 \cdot 1}}\right ) \\ & = \arctan \left (\frac {1}{F_2}\right ) \\ & = \arctan \left (1\right ) \\ & = \frac {\pi }{4}. \end {align*}