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Sum of reciprocal sequence

Let \{a_n\}_{n \in \mathbb{N}} be a sequence such that a_1=3 and

    \[a_{n+1} = a^2_{n}-2  \quad \text{forall} \;\; n \geq 2\]

Evaluate the sum

    \[\mathcal{S}  = \sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{1}{a_k}\]

Solution

Let us consider the sequence

    \[b_n=\frac{a_n-\sqrt{a_n^2-4}}{2}\]

and observe that

    \begin{align*} b_{n+1}&=\frac{a_{n+1}-\sqrt{a_{n+1}^2-4}}{2} \\ &=\frac{a_{n}^2-2-\sqrt{(a_{n}^2-2)^2-4}}{2} \\ &=\frac{a_{n} \left(a_n-\sqrt{a_n^2-4} \right)-2}{2} \\ &=a_nb_{n}-1 \end{align*}

This in return means,

    \[b_n=\frac{1}{a_n}+\frac{b_{n+1}}{a_n}\]

Thus,

    \begin{align*} b_1 &= \frac{1}{a_1} + \frac{b_2}{a_1} \\ & = \frac{1}{a_1} + \frac{1}{a_1} \left ( \frac{1}{a_2} + \frac{b_3}{a_2} \right ) \\ & = \frac{1}{a_1} + \frac{1}{a_1 a_2} \left ( \frac{1}{a_3} + \frac{b_4}{a_3} \right ) \\ &= \cdots \end{align*}

which is the desired result. Therefore,

    \[\mathcal{S} = \sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{1}{y_k} = b_1 = \frac{3-\sqrt{5}}{2}\]

Notes:

  1. The recursive relation a_{n+1} = a_n^2 + c has a closed form if-f c=0 or c=-2. In our case it is:

        \[a_n = \alpha^{2^{n-1}}+\frac{1}{\alpha^{2^{n-1}}}\]

    where \alpha = \frac{3 \pm \sqrt{5}}{2}.

  2. The number 3-\sqrt{5} is known as the Grafting constant.
  3. Under the same assumptions it holds that

        \[\lim_{n \rightarrow +\infty} \left( a_1 a_2 a_3 \dots a_n \right)^{\frac{1}{2^{n}}}=\frac{3+\sqrt{5}}{2}\]

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