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Assume all terms $a(n) >0$ , $\sqrt{a(1)}\geqslant 1 +\sqrt{a(0)}$, and $$\left|\dfrac{a(n+1)}{a(n)}-\dfrac{a(n)}{a(n-1)}\right|\leqslant \dfrac{1}{a(n)} $$ for all $n>0$.

  1. Prove that $\dfrac{a(n+1)}{a(n)}$ is convergent to some $L\geqslant 1$.

  2. Prove that $L>1$.

  3. Prove that $\dfrac{a(n)}{L^n}$ is convergent.

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sorry but where did L come from? –  kaine Jan 18 '13 at 21:00
    
It's hiding in the first part: $\displaystyle L = \lim_{n\to\infty} \frac{a(n+1)}{a(n)}$. –  Eric Stucky Jan 18 '13 at 21:01
    
Does "then" actually mean that the condition must hold? Because otherwise, the inequality recurrence is true for the sequence $a(i) = 2^{2^i}$, but none of the rest need to hold. –  Calvin Lin Jan 18 '13 at 21:05
    
Maybe “then” means: For any $n>0$, the inequality holds. (I.e., if $n>0$, then …) –  Harald Hanche-Olsen Jan 18 '13 at 21:14
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As @user58811 does not seem interested in clearing up the confusion about the meaning, I have taken the liberty of doing a possibly harmful edit. “In order to save the question, I had to destroy it.” user58811 should change it if I got it wrong. –  Harald Hanche-Olsen Jan 18 '13 at 21:32
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