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I tried attempting the question, and the best upper bound I could obtain was $1+\ln{98}$. I tried using $A_{n}\le n$ to form a harmonic series, but that wasn't strong enough. Any help would be appreciated, Thanks.

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  • $\begingroup$ Actually, $A_{97} > 14$, from a computation. $\endgroup$ – lhf Jun 25 '15 at 11:43
  • $\begingroup$ Seeking all functions $f:\mathbb{N}\to \mathbb{R}$ satisfy the functional equation $f^2(n-1)-f(n)f(n-1)+1=0$, with $f(1)=1$. Try to solve this eq. $\endgroup$ – Mohammad W. Alomari Jun 25 '15 at 12:11
  • $\begingroup$ @mwomath No thanks (unless you get MUCH more specific). $\endgroup$ – Did Jun 25 '15 at 12:55
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HINT: Try squaring your recursion.

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Just to spell out Michael's answer:

Let $B_n = A_n^2 + \frac{1}{A_n^2}$. Squaring the recursive relation, you find $B_n > A_n^2 = B_{n-1} + 2$. Hence, $B_{100} > B_{99} + 2 > \dots > B_1 + 198 = 200$. It follows that $A_{100}^2 + \frac{1}{A_{100}^2} > 200$. Solving, as you can, equation $x^2 + \frac{1}{x^2} = 200$, and using that $A_{100} > 1$, you find that $$A_{100} > \sqrt{100+3 \sqrt{1111}} = 14.142\dots > 14,$$ which is what you needed.

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Another fleshing out of Michael's answer:

Let $B_n = A_n^2$. Then $B_n=A_{n-1}^2 + 2 + \frac{1}{A_{n-1}^2} > B_{n-1} + 2$.

Thus, $B_n > B_1 + (n-1)2=2n-1$.

Now, $A_n > 14$ iff $B_n > 196$, and so it suffices to take $2n-1>196$, that is, $n \ge 99$.

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  • $\begingroup$ Perhaps more interesting is proving that $A_n \sim \sqrt{2n}$. $\endgroup$ – lhf Jun 26 '15 at 2:13
  • $\begingroup$ Or $A_n\sim\sqrt{2n+k\ln n}$, I can't work out $k$ at the moment. $\endgroup$ – Empy2 Jun 26 '15 at 17:09
  • $\begingroup$ @Michael $$k=\frac12$$ $\endgroup$ – Did Jun 28 '15 at 9:04

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