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I tried to do this using the $y=\frac{1}{x}$ hyperbola, didn't get anywhere though, I did find other things though like $\ln(2)>\frac{1}{2}$ and $\ln(3)-\ln(2)<\frac{1}{2}$.

stange
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Paradox
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    The statement is the same as saying $\int_1^3 \frac{dx}{x} > 1$. You can try to bound the integral from below by e.g. dividing $[1,3]$ into $N$ equal pieces and doing the sum ($N=7$ should do... and I would use a computer to do the sum). – Winther Oct 27 '23 at 10:12
  • The answer depends on how you are defining the function $\ln$. Is it defined as the inverse of the map $x\mapsto e^x$, and if so, how are you defining $e^x$? Or perhaps you are using the definition $\ln x = \int_{1}^{x}\frac{1}{t} , dt$? It would help if you edit your question for clarity. – Joe Oct 27 '23 at 10:16
  • I'm defining ln(x) as the integral you proposed – Paradox Oct 27 '23 at 10:27
  • I'm also defining ln(x) as the inverse. Here's how I define e^x: For all solutions to y'=y the solutions are y=ce^x where c is a constant. From this follows the reciprocal of the factorials infinite summation for Euler's number and the fact Euler's number is the limit as x approaches infinity of (1+1/x)^x. – Paradox Oct 27 '23 at 10:42

6 Answers6

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Applying the Hermite-Hadamard inequality (see also here) to the strictly convex function $x \mapsto 1/x$ gives the lower bound $$ \ln (3) = \int_1^3 \frac{dx}{x} > \frac{2}{(1+3)/2} = 1 \, . $$

This inequality simply uses the fact that the graph of a convex function lies above any of its tangents, here the tangent at the point $(2, 1/2)$:

enter image description here

Martin R
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Yes, just divide the axis between 1 and 3 into 2n intervals of width 1/n. Since 1/t is decreasing, you get $1/t \geq 1/(1+k/n)$ on the interval $[1+(k-1)/n, 1+k/n]$ and thus $$ \ln(3) = \int_1^3 {dt \over t} \geq {1 \over n}\sum_{k=1}^{2n} {1\over 1+{k\over n}} = \sum_{k=1}^{2n} {1\over n+k}. $$ For n = 4, the right sum is already bigger than 1. It is equal to 28721/27720.

coudy
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Using the Taylor series of the logarithm we get for $0 < x < 1$ $$ \ln \left( \frac{1+x}{1-x}\right) = \ln(1+x) -\ln(1-x) = 2 \left( x + \frac 13 x^3 + \frac 15 x^5 + \cdots \right) > 2x \, . $$ Setting $x=1/2$ gives $\log(3) > 1$.

Martin R
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  • (+1) Hello @MartinR, I have an "elementary" proof using $e^x≥x+1$ . But I am not an expert in calculus, is this argument circular ? – lone student Oct 27 '23 at 13:30
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    @lonestudent: I don't know, it probably depends on how you show that $e^x \ge 1+x$, there are many proofs of that inequality: https://math.stackexchange.com/q/504663/42969 – Martin R Oct 27 '23 at 13:38
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    Good or bad, I have posted the proof. It's a enough simple proof that can be done by anyone . – lone student Oct 27 '23 at 18:42
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We can also reach the proof by simply applying the well-known inequality $e^x≥x+1$ .

Using the substitution $x\longmapsto -\ln x,\thinspace x>0$ we have :

$$ \begin{align}&\frac 1x≥1-\ln x\\ \iff &\ln x≥1-\frac 1x \color{#c00}{\tag 1}\end{align} $$

Equality occurs iff, when $x=1$ .

Then, we see that putting $x=3$ leads to :

$$ \begin{align}\ln 3>1-\frac 13=\frac 23\end{align} $$

which is too weak .

Therefore, we want to multiply both side of the inequality $\color{#c00}{(1)}$ by $6$ and see if it works for us :

$$ \begin{align}&\ln x^6≥6-\frac 6x,\thinspace x>0\\ \iff &\ln x \overset{x\thinspace\longmapsto\thinspace\sqrt[6]{x}}{≥}6-\frac {6}{\sqrt [6]{x}}\end{align} $$

Thus, it is enough to show that :

$$6-\frac {6}{\sqrt [6]{3}}>1$$

which is equivalent to :

$$ \begin{align}&3>\left(\frac 65\right)^6\\ \iff &5^6>2\cdot 6^5\\ \iff &15625>15552 \thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

lone student
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Use the Taylor series for $\ln(x)$ at $x=1$:

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}\mp...$$

The radius of convergence is $|x|<1$, so $x=2$ won't do. However, take $x=-\frac{2}{3}$, then

$$\ln\left(\frac{1}{3}\right)=-\frac{2}{3}-\frac{2}{9}-\frac{8}{81}-\frac{4}{81}-...<-1$$

where the first four terms are enough to dip below $-1$. Therefore, $\ln(3)=-\ln\left(\frac{1}{3}\right)>1$.

If you are starting from the integral definition for the natural logarithm, you can of course derive the Taylor series around $1$ by integrating the geometric series $\frac{1}{1+x}=\sum_{n\ge 0} x^n$.

lone student
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Joshua Tilley
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$$\begin{align} \ln(3) &= \frac{⅔^1}{1} + \frac{⅔^2}{2} + \frac{⅔^3}{3} + \frac{⅔^4}{4} + ⋯\\ &> \frac{⅔^1}{1} + \frac{⅔^2}{2} + \frac{⅔^3}{3} + \frac{⅔^4}{4}\\ &= 1.037037037037⋯\\ &> 1. \end{align}$$

NinjaDarth
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