<s>[INST] How to show that $\ln(2) > \frac{2}{3}$ I'm trying to show that $\ln(2) > \frac{2}{3}$. However, I'm pretty stuck on how to proceed.
I tried to show that this is the case using a similar technique to
$$
e < 4 \iff \sqrt{e} < 2 \iff \frac{1}{2} < \ln(2)
$$
However, while I can show that $\sqrt{e} < 2$ (because $2^2 = 4 > e$), I don't think it's easy to do this when $\frac{2}{3}$ is the exponent (at least it's not easy for me).
The other approach I can think of is to use a series expansion for $\ln(2)$, but the only one I know is $\ln(1+x)$ for $\lvert x\rvert < 1$.
 [/INST] $\ln(2) > \frac{2}{3}$ is equivalent to $e^2 < 8$, so we need an upper bound for $e$. A standard trick is to replace the “tail” of the exponential series by a geometric series, which can be computed explicitly:
$$
 e = \sum_{n=0}^\infty \frac{1}{n!}= 1 + 1 + \frac 12  \left(
1 + \frac{1}{3} + \frac{1}{3\cdot 4} + \cdots\right) \\
< 1 + 1 + \frac 12 \left(
1 + \frac{1}{3} + \frac{1}{3^2} + \cdots\right) \\
= 2 + \frac 12 \cdot \frac {1}{1-1/3} = \frac{11}{4} \, .
$$
It follows that
$$
 e^2 < \left(\frac{11}{4} \right)^2 = \frac{121}{16} < 8
$$
and taking logarithms we get $2/3 < \ln(2)$.
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