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Let $\langle a_n \rangle$ be a recursive sequence given by $a_1>2$ and,

$a_{n+1}=a_n^2-2$ for $n \in \mathbb N$

Show that

$\sum_{n=1}^\infty \frac{1}{a_1a_2\cdots a_n} = \frac{a_1-\sqrt{a_1^2-4}}{2}$

I have reached this step:

$\frac{1}{a_1a_2a_3\cdots a_n}=\frac{1}{2} (\frac{a_n}{a_1a_2\cdots a_n-1} -\frac{a_n-1}{a_1a_2\cdots a_n})$

But I am not able to obtain the final expression .Please help me to obtain it. Thanks for help in advance.

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    $\begingroup$ The required fraction looks like the quadratic formula. $\endgroup$
    – Idonknow
    Commented Aug 19, 2017 at 13:02
  • $\begingroup$ @please use the math I provided and get rid of the image. $\endgroup$
    – jimjim
    Commented Aug 19, 2017 at 13:58

2 Answers 2

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Notice the function $x \mapsto x^2 - 2$ send $(2,\infty)$ to $(2,\infty)$.

Start from $a_1 > 2$, it is easy to see $a_n > 2$ for all $n$.

For each $n$, pick a $x_n > 1$ such that $a_n = x_n + \frac{1}{x_n}$, the recurrence relation becomes

$$x_{n+1} + \frac{1}{x_{n+1}} = \left(x_n + \frac{1}{x_n}\right)^2 - 2 = x_{n}^2 + \frac{1}{x_{n}^2}\quad\implies\quad x_{n+1} = x_n^2$$

Let $\lambda = x_1$. Solving above recurrence relation gives us $$x_n = \lambda^{2^{n-1}}\quad \implies\quad a_n = \lambda^{2^{n-1}} + \lambda^{-2^{n-1}} = \frac{\lambda^{2^n} - \lambda^{-2^n}}{\lambda^{2^{n-1}} - \lambda^{-2^{n-1}}} $$ The summands in the sum are telescoping products.

$$\begin{align}\prod_{k=1}^n \frac{1}{a_k} &= \prod_{k=1}^n \frac{\lambda^{2^{k-1}} - \lambda^{-2^{k-1}}}{\lambda^{2^{k}} - \lambda^{-2^k}} = \frac{\lambda - \lambda^{-1}}{\lambda^{2^n} - \lambda^{-2^n}}\\ &= \frac{\lambda^2-1}{\lambda}\frac{\lambda^{2^n}}{(\lambda^{2^n})^2 - 1} = \frac{\lambda^2-1}{\lambda}\left[\frac{1}{\lambda^{2^n}-1} - \frac{1}{\lambda^{2^{n+1}} - 1}\right] \end{align} $$

The sum itself is also a telescoping one. At the end, we have

$$\begin{align}\sum_{n=1}^\infty \prod_{k=1}^n \frac{1}{a_k} &= \frac{\lambda^2-1}{\lambda} \sum_{n=1}^\infty \left[\frac{1}{\lambda^{2^n}-1} - \frac{1}{\lambda^{2^{n+1}} - 1}\right] = \frac{1}{\lambda}\\ &= \frac12\left( (\lambda + \lambda^{-1}) - (\lambda - \lambda^{-1}) \right) = \frac12\left( (\lambda + \lambda^{-1}) - \sqrt{(\lambda + \lambda^{-1})^2 - 4} \right)\\ &= \frac{a_1 - \sqrt{a_1^2-4}}{2} \end{align} $$

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  • $\begingroup$ How to obtain the very last equality? $\endgroup$
    – Idonknow
    Commented Aug 20, 2017 at 0:57
  • $\begingroup$ @Idonknow I thought that algebraic trick is standard, see update. $\endgroup$ Commented Aug 20, 2017 at 1:04
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Let $P_n = a_1 a_2 \cdots a_n$, and let $b_n = a_{n+1}/P_n$.

From the recursion, note that $$a_{n+1} - 2 = {a_n}^2 - 4 \implies a_n + 2 = \frac{a_{n+1} - 2}{a_n - 2},$$ hence we have the telescoping product $${P_n}^2 = \prod_{k=2}^{n+1} (a_k+2) = \prod_{k=2}^{n+1} \frac{a_{k+1} - 2}{a_k - 2} = \frac{a_{n+2} - 2}{a_2 - 2}$$

It follows that $${b_n}^2 = (a_2 - 2) \frac{{a_{n+1}}^2}{a_{n+2} - 2} = (a_2 - 2) \frac{a_{n+2} + 2}{a_{n+2} - 2} \implies \lim_{n\to\infty} b_n = \sqrt{a_2 - 2} = \sqrt{{a_1}^2 - 4}$$

Next, note that $$b_{k-1} - b_{k} = \frac{{a_k}^2}{P_k} - \frac{a_{k+1}}{P_k} = \frac{2}{P_n} \implies 2\sum_{k=2}^{n} \frac{1}{P_k} = b_1 - b_n $$ $$\implies 2\sum_{k=1}^{n} \frac{1}{P_k} = \frac{a_2+2}{a_1} - b_n = a_1 - b_n$$

We conclude that $$\sum_{k=1}^{\infty} \frac{1}{P_k} = \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{P_k} = \lim_{n\to\infty} \frac{a_1 - b_n}{2} = \frac{a_1 - \sqrt{{a_1}^2 - 4}}{2}$$

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