What is reasoning behind statement that $\log \frac{3}{2} < (\frac{3}{2} - 1)$?

Question

In his paper, "Euler and the Partial Sums of the Prime Harmonic Series" (pdf), Paul Pollacks makes a small statement at the top of page 4:

$$ \log \frac{3}{2} < (\frac{3}{2} - 1) = \frac12 $$

I can see numerically this is true, but I can't see how he makes the inequality $\log \frac{3}{2} < (\frac{3}{2} - 1)$.

I have looked through references of power series approximations but can't see anything that matches.

See here for a proof of the general case. – user3482749 Mar 21 '21 at 23:49 — user3482749, Mar 21 '21 at 23:49
usually written $\log (1+x) < x$ – Will Jagy Mar 21 '21 at 23:51 — Will Jagy, Mar 21 '21 at 23:51

hunter · Accepted Answer · 2021-03-21T23:50:27.850

2

For any $x > 1$ we have $$ \log(x) < x - 1. $$ For a geometric proof, note that both functions vanish at $x = 1$ and that the graph of $\log(x)$ is concave down (so is trapped below the tangent line as you move to the right).

edited Mar 21 '21 at 23:50

answered Mar 21 '21 at 23:48

hunter

29,847

you seem to have the inequality sign backwards – Will Jagy Mar 21 '21 at 23:50
@WillJagy fixed, thanks! – hunter Mar 21 '21 at 23:50
@hunter is there an argument to be made using the gradient of $\log(x)=1/x$ to formalise the statement that $y=\log(x)$ is always less than $y=x-1$ ? – Penelope Mar 21 '21 at 23:55
1

@tariqrashid the first derivative gives the slope of the tangent line and the second derivative being negative shows that the graph is concave down. – Karl Mar 22 '21 at 00:14

score 1 · Answer 2 · answered Mar 22 '21 at 00:55

If $x = 1$ then $0 =\ln x = x-1$.

Would you accept that as $x, x-1$ and $\ln x$ get larger than $x$ and $x-1$ getter larger "faster" than $x$ does?

We can argue that with derivatives.

$\frac {dx}{dx} = \frac {d(x-1)}{dx} = 1$ but for $x > 1$ then $\frac {d\ln x}{dx} = \frac 1x < 1$.

So $\ln x$ increases more "slowly" than $x-1$.

So if they start on the racetrack at the same value at $x = 1$ (so $x-1 =0$ and $\ln x = 0$) but $x-1$ is always faster than $\ln x$ which just gets slower and slower, then $x-1$ immediate surpasses it.

Of course.... I have to take my roommate to the hospital as I just knocked his eye out but waving my hands about so wildly.)

score 0 · Answer 3 · answered Mar 22 '21 at 00:16

0

You can easily check that the function

$$f(x)= x-1-\log x$$

has a minimum equal to $0$ at $x=1$ through derivative's test , so for all other points $f(x) >0$

answered Mar 22 '21 at 00:16

Tortar

3,980

What is reasoning behind statement that $\log \frac{3}{2} < (\frac{3}{2} - 1)$?

3 Answers3