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I have to prove this inequality, where $Q_k$ is the $k$-th Fibonacci number and $R_k$ is the $k$-th convergent of $[1;1,1,...]$:

$$R_1=1$$ $$R_2=1+\frac{1}{1}$$ $$R_3=1+\frac{1}{1+\frac{1}{1}}$$

and so on.

So, from the following properties of the sequence $\{R_k\}$:

$$R_k=\frac{Q_{k+1}}{Q_k}$$ $$R_{k+1}-R_k=\frac{(-1)^k}{Q_{k+1}Q_k}$$ $$Q_{k+2}=Q_{k+1}+Q_k$$

I found that the Fibonacci sequence can be written taking into account only one previous term:

$$Q_{k+1}=\frac{Q_{k}}{2}\left(1+\sqrt{5-\frac{4(-1)^k}{Q_{k}^2}}\right)$$

and so in this way I can obtain the exact expression for the error between $R_k$ and its limit value:

$$E_k=\left | \frac{1+\sqrt{5}}{2}-\frac{Q_{k+1}}{Q_k} \right | =\frac{(-1)^{k+1}}{2}\left( \sqrt{5}-\sqrt{5+\frac{4(-1)^k}{Q_k^2}} \right).$$

Now, if I try to prove that this last expression verify the claim in the case of $k$ even, I obtain:

$$-\frac{1}{2}\left( \sqrt{5}-\sqrt{5+\frac{4}{Q_k^2}}\right) >\frac{1}{Q_k^2\sqrt{5}} \Rightarrow 0>\frac{4}{Q_k^4}$$

that is an absurd.

Where am I doing wrong?

Nameless
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  • Can you show us how you got $$0>\frac{4}{Q_k^4}$$ explicitly? That is, please expand the last result by quite a bit. – Batominovski Aug 11 '18 at 10:26
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    Plus, I also suspect that the inequality is true only when $k$ is an odd positive integer. – Batominovski Aug 11 '18 at 10:37
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    To support @Batominovski's argument https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) and https://math.stackexchange.com/questions/2058057/prove-the-inequality-left-fracmn-frac1-sqrt52-right-frac1mn/ – rtybase Aug 11 '18 at 11:09

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You have $$E_k=\frac{\sqrt{5}}{2}\,\left|\left(1+\frac{4\,(-1)^k}{5\,Q_k^2}\right)^{\frac{1}{2}}-1\right|\text{ for }k=1,2,3,\ldots\,.$$ Using Bernoulli's Inequality, we see that $$\left(1+\frac{4\,(-1)^k}{5\,Q_k^2}\right)^{\frac12}< 1+\frac{2\,(-1)^k}{5\,Q_k^2}\text{ for }k=1,2,3,\ldots\,.\tag{*}$$ This shows that, if $k\in\mathbb{Z}_{>0}$ is odd, then $$E_k=\frac{\sqrt{5}}{2}\,\left(1-\left(1-\frac{4}{5\,Q_k^2}\right)^{\frac12}\right)>\frac{\sqrt{5}}{2}\,\left(\frac{2}{5\,Q_k^2}\right)=\frac{1}{\sqrt{5}\,Q_k^2}\,,$$ as required. However, if $k\in\mathbb{Z}_{>0}$ is even, then (*) proves that $$E_k=\frac{\sqrt{5}}{2}\,\left(\left(1+\frac{4}{5\,Q_k^2}\right)^{\frac12}-1\right)<\frac{\sqrt{5}}{2}\,\left(\frac{2}{5\,Q_k^2}\right)=\frac{1}{\sqrt{5}\,Q_k^2}\,.$$ Hence, what you want to prove is not entirely correct, and WolframAlpha confirmed my calculations for $k=1,2,3,4,5$: $$T_1\approx 1.38\,,\,\,T_2\approx 0.854\,,\,\,T_3\approx1.06\,,\,\,T_4\approx0.979\,,\text{ and }T_5\approx 1.01\,,$$ where $$T_k:=\sqrt{5}\,Q_k^2\,E_k\text{ for }k=1,2,3,\ldots\,.$$

Batominovski
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