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Lucas arctan series

Let L_n denote the n -th Lucas number, defined by L_0=2 , L_1=1 and for all n \geq 2

    \[L_{n} =L_{n+1} + L_{n+2}\]

Compute the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \arctan \left ( \frac{L_{n+1}^2}{1+L_n L_{n+1}^{2}L_{n+2}} \right )\]

Solution

Let \displaystyle a_n=\arctan \left ( \frac{L_{n+1}^2}{1+L_n L_{n+1}^{2}L_{n+2}} \right ). Thus,

    \begin{align*} a_n &= \arctan \left ( \frac{L_{n+1}^2}{1+L_n L_{n+1}^{2}L_{n+2}} \right ) \\ &= \arctan \left ( \frac{\frac{1}{L_n L_{n+2}}}{1+\frac{1}{L_n L_{n+1}}\cdot \frac{1}{L_{n+1} L_{n+2}}} \right ) \\ &= \arctan \left ( \frac{\frac{1}{L_n L_{n+1}}- \frac{1}{L_{n+1}L_{n+2}}}{1+\frac{1}{L_n L_{n+1}}\cdot L_{n+2}} \right )\\ &=\arctan \left ( \frac{1}{L_n L_{n+1}} \right ) - \arctan \left ( \frac{1}{L_{n+1}L_{n+2}} \right ) \end{align*}

Hence the series telescopes and thus \mathcal{S}=\arctan \frac{1}{3}.

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