This is just a minor extension of my last post.

Why is $$1/\log(1+1/k) \approx k + 1/2\ ?$$

At first, I was surprised that $$1/\log(1+1/237) \approx 237,$$ but then I realize that $$\log(1+x) \approx x$$ if $|x|$ is small, so

$$1/\log(1+x) \approx 1/x,\mathrm{\ thus}$$

$$1/\log(1+1/k) \approx k.$$

$$1/\log(1+1/k) \approx k.$$

Where does the 1/2 come from? Well, you’ve got to get a better approximation of $\log(1 +x)$ to find the 1/2. You can do this with calculus. If $|x| < $1 and $|y| < 1$, then $$(1-x)\cdot( 1+ x + x^2 + x^3 + \ldots) = 1$$

$$1/(1-x) = 1+ x + x^2 + x^3 + \ldots$$

$$\int_0^y 1/(1-x) \;dy = y + y^2/2 + y^3/3 + \ldots$$

$$-\log(1-y) = y + y^2/2 + y^3/3 + \ldots,\mathrm{\, so}$$

$$\log(1+x) = x – x^2/2 + O(x^3)$$ using big O notation. (Aside: $-\log( 1 – .001) = 0.0010005003335835335…$)

$$1/(1-x) = 1+ x + x^2 + x^3 + \ldots$$

$$\int_0^y 1/(1-x) \;dy = y + y^2/2 + y^3/3 + \ldots$$

$$-\log(1-y) = y + y^2/2 + y^3/3 + \ldots,\mathrm{\, so}$$

$$\log(1+x) = x – x^2/2 + O(x^3)$$ using big O notation. (Aside: $-\log( 1 – .001) = 0.0010005003335835335…$)

Now,

$$\begin{aligned} 1/\log(1+x)

&= 1/(x – x^2/2 + O(x^3)) \\

&= 1/x \cdot 1/(1 – x/2 + O(x^2)) \\

&= 1/x \cdot [ 1 + ( x/2 + O(x^2)) + O( ( x/2 + O(x^2))^2) ] \\

&= 1/x \cdot ( 1 + x/2 + O(x^2) + O( x^2) ) \\

&= 1/x \cdot ( 1 + x/2 + O( x^2) ) \\

&= 1/x + 1/2 + O(x).

\end{aligned}$$Replacing $x$ with $1/k$ gives$$1/\log(1+1/k) = k + 1/2 + O(1/k).$$

$$\begin{aligned} 1/\log(1+x)

&= 1/(x – x^2/2 + O(x^3)) \\

&= 1/x \cdot 1/(1 – x/2 + O(x^2)) \\

&= 1/x \cdot [ 1 + ( x/2 + O(x^2)) + O( ( x/2 + O(x^2))^2) ] \\

&= 1/x \cdot ( 1 + x/2 + O(x^2) + O( x^2) ) \\

&= 1/x \cdot ( 1 + x/2 + O( x^2) ) \\

&= 1/x + 1/2 + O(x).

\end{aligned}$$Replacing $x$ with $1/k$ gives$$1/\log(1+1/k) = k + 1/2 + O(1/k).$$

It is more work to prove sharper bounds.