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I would like to prove that $\exists n_0$ such that the sequence $a_n = \arctan\Big(\frac{100 \log^2 n}{\sqrt{n+1} - \sqrt[3]{n}}\Big)$ is decreasing $\forall n \ge n_0$. It is sufficient to show that the sequence $b_n = \frac{\log^2 n}{\sqrt{n+1} - \sqrt[3]{n}}$ is decreasing, the rest follows. I am trying to avoid derivatives.

I have taken the limit $\lim_{n \to\infty} \frac{b_n}{b_{n+1}} = 1$, getting that $\forall \epsilon > 0$ $\exists n_0$ $\forall n \ge n_0: $ $b_{n+1} < (1 + \epsilon)b_n$, but that does not prove anything. I have also been looking for bounds of $\log n$ without success.

Does anyone know some simple way of proving this? And, by the way - is there a tight bound for $\log n$?

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$\ln n$ is more or less $\sum_{i=1}^n 1/i$ - google for exact bounds. – savick01 Aug 21 '13 at 9:21
What do you mean by $\log^2n$ is it $(\log n)^2$? – savick01 Aug 21 '13 at 9:25
Yes, $\log^2 n = (\log n)^2$ where $\log n$ is the natural logarithm. – David Čepelík Aug 21 '13 at 9:27
Do you know any derivatives? Or convexity? If you allow me to estimate square root and logarithm near $1$ by affine functions (ideally by Taylor thm, but I suppose you don't know it, so am looking for some weaker tools), I think I know how to prove it. – savick01 Aug 21 '13 at 9:32
Please, use $\displaystyle{\ln\left(n\right)}$ instead of $\displaystyle{\log\left(n\right)}$. – Felix Marin Sep 5 '13 at 23:28

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