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 $\lvert x\rvert < 1$.
 A: You can define $\ln 2$ by $ \displaystyle \ln 2 = \int_1^2 \frac 1t \, dt$. Any Riemann sum for this integral with right endpoints will be less than $\ln 2$ because $\dfrac 1t$ is decreasing. In particular, if you subdivide $[1,2]$ into $n$ equal subintervals you obtain
$$\ln 2 > \frac 1{n+1} + \cdots + \frac 1{2n}.$$
You may need to experiment a bit with different values of $n$, but before $n$ gets too large you will obtain $\ln 2 > \dfrac 23$.
A: Indeed, considering series like
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...\tag{1}$$
can be interesting but this one has a too slow rate of convergence.
Instead use this one (derived from (1)):
$$\ln\left(\frac{1+x}{1-x}\right)=2x+\frac23x^3+\frac25x^5+...\tag{2}$$
in which it suffices to take $x=\frac13$ to get:
$$\ln(2)=\frac23+\text{a positive quantity}$$
establishing the claim.
Remark:
Series (2) has been used to compute certain values of logarithmic tables before the advent of computers.
A: $\ln(2) > \frac{2}{3}$ 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 $2/3 < \ln(2)$.
