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Can it be shown that

\begin{align} \frac{1}{\ln(1+L_{n}) -1} \geq \frac{L_{n}}{(L_{n}-1)(e^{L_{n}}-1)} \end{align}

where $L_{n}$ is the $n^{th}$ Lucas number. Show results in full detail.

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  • $\begingroup$ Is this really the inequality you're looking for? The $e^{L_n}$ term increases so fast that the result clearly holds for large $L_n$; you just need to verify it for the first few terms. $\endgroup$ – mjqxxxx May 14 '14 at 1:23
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    $\begingroup$ "Show results in full detail." I wish Mariano were around to explain the impoliteness of the imperative. $\endgroup$ – Cheerful Parsnip May 14 '14 at 1:24
  • $\begingroup$ Often math problems do start with comparing the first few values, but this only lasts until someone asks for a proof. Upon this request of proof, most, including me, often say "uh-oh" how to prove this one as well. $\endgroup$ – Leucippus May 14 '14 at 1:31
  • $\begingroup$ @Leucippus: have you tried using power series? $\endgroup$ – DeepSea May 14 '14 at 1:43
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    $\begingroup$ @Leucippus: The way it is phrased seems like an order. More polite would be "I'm interested in a complete proof." $\endgroup$ – Cheerful Parsnip May 14 '14 at 5:41
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$ln(1+L_n) \leq L_n \to ln(1 + L_n) - 1 \leq L_n - 1 \to \dfrac{1}{ln(1 + L_n) - 1} \geq \dfrac{1}{L_n - 1}$.

Also: $e^{L_n} \geq L_n + 1 \to e^{L_n} - 1 \geq L_n \to 1 \geq \dfrac{L_n}{e^{L_n} - 1}$.

The conclusion follows from these inequalities.

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  • $\begingroup$ nice solution, clean and fast $\endgroup$ – Leucippus May 14 '14 at 2:05

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