Estimating area below a hyperbola : showing $\log(3)\gt 1$

Question

In a recent answer of mine, I needed to find an explicit constant rational number $c$ such that $\log(3)\geq c \gt 1$ without using a calculator. I proceeded using a Riemann sum :

$$\log(3)=\int_{1}^{3}\frac{dt}{t}=\sum_{k=1}^8 \int_{1+\frac{k-1}{4}}^{1+\frac{k}{4}}\frac{dt}{t}\geq \sum_{k=1}^8\frac{1}{4+k}=\frac{28271}{27720} $$

If we look at the complexity of the last step in this computation, we have a sum of $8$ rational numbers, so this makes $3\times 7$ multiplications, $7$ additions, and $7$ fraction reductions.

Since there are estimates with smaller numbers, such as $\log(3)\gt \frac{12}{11}$, I wonder if there are simpler (calculator-free) proofs.

Perhaps by taking a suitable tangent below the hyperbola to evaluate the area ?

Answer 1

$\forall y\in(0,+\infty)\quad\ln y=2\sum_{k=0}^{\infty}\frac1{2k+1}\left(\frac{y-1}{y+1}\right)^{2k+1}$ hence $$\ln3>2\left(\frac12+\frac1{3\cdot2^3}\right)=\frac{13}{12}.$$

Answer 2

Solution 1 The Taylor series for $$\log \left(\frac{1+x}{1-x}\right) = 2 \operatorname{artanh} x$$ is $2\left(x + \frac{1}{3} x^3 + \frac{1}{5} x^5 + \cdots\right)$, with all coefficients nonnegative, so for positive $x$, $$2 \operatorname{artanh x} > 2 x + \frac{x^3}{3} ,$$ and in particular $$\log 3 = 2 \operatorname{artanh} \frac{1}{2} > 2 \left[ \frac{1}{2} + \frac{1}{3}\left(\frac{1}{2}\right)^3 \right] = \frac{13}{12} > 1 .$$

Solution 2 Alternatively, evaluating a Dalzell-style integral gives $$0 < \int_0^\sqrt{2} \frac{x^2 (1 - x)^2}{1 + x^2} = \log 3 - \left(2 - \frac{2}{3} \sqrt{2}\right),$$ so it suffices to find a rational number between $1$ and $2 - \frac{2}{3} \sqrt{2}$, or, by rearranging, a number between $2 \sqrt{2}$ and $3$, but rearranging shows that $\frac{17}{6}$ suffices for the former, and correspondingly $1 < \frac{19}{18} < \log 3$. Probably a more clever choice of integral can reduce the amount of subsequent computation.

Remark Incidentally, direct computation shows that the (signed) area over the interval $[1, 3]$ and under the tangent line to the curve $x y = 1$ at $(a, \frac{1}{a})$ is $\frac{4 (a - 1)}{a^2}$, which has a maximum of $1$ (at $a = 2$), hence without modification this method cannot yield an explicit rational between $1$ and $\log 3$.