Natural Logarithm of 2 is Greater than One Half

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Lemma

$\ln 2 \ge \dfrac 1 2$

where $\ln$ denotes the natural logarithm function.


Proof 1

Let $f: \R_{>0} \to \R$ be the real function defined as:

$\forall x \in \R_{>0}: f \left({x}\right) = \dfrac 1 x$

From Real Rational Function is Continuous, $f \left({x}\right)$ is a continuous real function, in particular on the closed interval $\left[{a \,.\,.\, b}\right]$.


Hence the Mean Value Theorem for Integrals can be applied:

There exists some $k \in \left[{1 \,.\,.\, 2}\right]$ such that:

\(\displaystyle \int_1^2 f \left({x}\right) \ \mathrm d x\) \(=\) \(\displaystyle f \left({k}\right) \left({2 - 1}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle \int_1^2 \frac 1 x \ \mathrm d x\) \(=\) \(\displaystyle \frac 1 k \left({2 - 1}\right)\) substituting $\dfrac 1 x$ for $f \left({x}\right)$
\(\displaystyle \implies \ \ \) \(\displaystyle \ln 2 - \ln 1\) \(=\) \(\displaystyle \frac 1 k \left({2 - 1}\right)\) Definition of Natural Logarithm
\(\displaystyle \implies \ \ \) \(\displaystyle \ln 2\) \(=\) \(\displaystyle \frac 1 k\) Logarithm of $1$ is $0$


Thus:

\(\displaystyle 1\) \(\le\) \(\displaystyle k \le 2\)
\(\displaystyle \implies \ \ \) \(\displaystyle 1^{-1}\) \(\ge\) \(\displaystyle k^{-1} \ge 2^{-1}\) Ordering of Reciprocals
\(\displaystyle \implies \ \ \) \(\displaystyle 1\) \(\ge\) \(\displaystyle \ln 2 \ge \frac 1 2\)

$\blacksquare$


Proof 2

\(\displaystyle 1 - \frac 1 x\) \(\le\) \(\displaystyle \ln x\) Lower Bound of Natural Logarithm
\(\displaystyle \implies \ \ \) \(\displaystyle \frac 1 2\) \(\le\) \(\displaystyle \ln 2\) letting $x = 2$

$\blacksquare$


Also see


Sources