What is reasoning behind statement that $\log \frac{3}{2} < (\frac{3}{2} - 1)$? 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.
 A: 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).
A: 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.)
A: 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 $f(x) >0$
