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In this answer to a question on CS, I claimed that the statement referred to in my question is true; thinking about it more clearly, I find that I can't prove that that is the case, nor do I know any reason why it would be other than "looks like it when I plot it on wolfram". I tried to bound it from below by some non-negative function without making any progress.

Context: the reason I used this statement was to prove that $f: \mathbb{R}_+ \to \mathbb{R}, f(x) := \left(1 + \frac{\log 2}{\log x}\right)^{\log x}$ is monotonically nondecreasing in order to use the series expansion for polynomials, which I only know to be valid for $\log x \in \mathbb{N}$. Given that that's not the case for all naturals $x$, I tried to use the monotonicity of $f$ so I can use $\lceil\log x\rceil$ instead.

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2 Answers 2

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Yes. Use, for $$ 0 \leq t \leq 1, \; \; \; \log(1+t) \geq \; t - \frac{t^2}{2} $$ which works because, with $0 \leq t \leq 1,$ the power series for $\log(1+t)$ is strictly alternating, with the absolute values of the terms decreasing.

Taking $$ t = \frac{\log 2}{\log x} $$ and $x > 2$ works out.

Oh, you need $\log (2x) = \log 2 + \log x.$

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  • $\begingroup$ Thanks a lot, that works out nicely; now I know I didn't give a wrong answer. $\endgroup$
    – G. Bach
    Aug 30, 2013 at 5:37
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As an aside, here is a simple proof that $\ln(1+t) \ge t-\dfrac{t^2}{2} $ for $0 \le t \le 1$.

All that is needed is $\dfrac{1}{1+t} \ge 1-t$ for $0 \le t \le 1$, and this follows from $1 \ge (1+t)(1-t) = 1-t^2 $.

$\begin{align} \ln(1+t) &= \int_1^{1+t} \dfrac{dx}{x}\\ &= \int_0^{t} \dfrac{dx}{1+x}\\ &\ge \int_0^{t} (1-x)\ dx\\ &= (x-\dfrac{x^2}{2})\big|_0^t\\ &= t-\dfrac{t^2}{2}\\ \end{align} $

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