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Let $\{b_{n}\}$ be a sequence defined as

$b_{n+1}=b^2_{n}-2$ and $b_{1}=b,$ where $b>2$.

Then find sum of $$\sum^{\infty}_{n=1}\frac{1}{b_{1}b_{2}b_{3}\cdots b_{n}}=$$

What i try ::

From equation $b_{n+1}=b^2_{n}-2$. Then $b_{2}=b^2-2$

And $b_{3}=b^2_{2}-2=(b^2-2)^2$

Now when i find for $b_{4},b_{5},b_{6},..$. Then expression have more terms and higher power.

I did not understand how can i find an

Less complex way to solve it.

Help me please. Thanks

jacky
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    See "solution 1" here found using https://approach0.xyz/ that I recommand you – Jean Marie Aug 15 '20 at 11:52
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    You have yourself found the solution, $(x^2-2)$ starts from a lower number just above 2, (so that we are not stuck at the sequence 2,2,2,2,2 for b=2)......It's easy to predict that as the index n goes higher we're left with a higher number. Do you see now? – Anindya Prithvi Aug 15 '20 at 12:45
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    $b^2_{2}-2\neq(b^2-2)^2$ – cr001 Aug 15 '20 at 13:26

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