Why are these two functions so close in value? [Edit: rewritten 4/8/22 to increase specificity of question]
Recently, I was interested to estimate the following partial sum.
$$\sum_{j=1}^{n-1}\frac{1}{\sqrt{j(j+1)}}$$
I understand that the correct asymptotic formula is $\ln n$, as $\frac1{\sqrt{j^2+j}} \sim \frac 1j$ for large $j$, and therefore
$$\sum_{j=1}^{n-1}\frac{1}{\sqrt{j(j+1)}} \sim \sum_{j=1}^{n-1}\frac1j \sim \ln n.$$
However, the error between the first and last terms is curiously much lower than you might expect, based on the error of the two intermediary approximations. By taking the forward difference on each side, one discovers the asymptotic formula
$$\ln\Big(\frac{x+1}{x}\Big) \sim \frac 1{\sqrt{x(x+1)}}$$
which is indeed remarkably close for large $x$. A partial explanation is: the power series expansion of each of these functions at $x = \infty$ begins with $\frac 1x - \frac1{2x^2} + \mathcal O \big(\frac1{x^3}\big)$. Indeed,
$$\frac 1{\sqrt{x(x+1)}} - \ln\Big(\frac{x+1}{x}\Big) = \frac1{24x^3} + \mathcal O\Big(\frac1{x^4}\Big)$$
However, this is somewhat unsatisfying to me as an explanation, as it seems to explain the coincidence with another coincidence.
As the comments have noted, it is easy to find examples of functions which more closely approximate $\frac{1}{\sqrt{x(x+1)}}$ for a similar reason. However, the above approximation is exceptionally striking as $\ln\big(\frac{x+1}{x}\big)$ may be written as the forward difference of $\ln x$, leading back to the originally observed partial sum approximation. Other functions, which may be closer to $\frac{1}{\sqrt{x(x+1)}}$ in value, are not so neatly (partially) summable.

Is there a deeper or more intuitive way to understand the above partial sum asymptote, without relying on the 'worse' intermediate approximation of $\frac{1}{\sqrt{x(x+1)}} \sim \frac 1x$?


Too long for a comment.
In the same spirit as @M. Wind's answer, we can build functions which are still closer.
Consider the three functions
$$f(x)=\log\Big(\frac{x+1}{x}\Big) \qquad \qquad g(x)=\frac 1{\sqrt{x(x+1)}}\qquad \qquad h(x)=\frac {12x}{12 x^2+6 x-1 }$$
$h(x)$ is a Padé approximant of $f(x)$.
As you wrote
$$f(x)- g(x)=-\frac{1}{24 x^3}+O\left(\frac{1}{x^4}\right) \qquad \text{but}\qquad f(x)- h(x)=-\frac{1}{24 x^4}+O\left(\frac{1}{x^5}\right)$$
Similarly
$$\Phi_1=\int_{10}^\infty \Big[f(x)-g(x)\Big]^2\,dx=2.73\times 10^{-9}$$
$$\Phi_2=\int_{10}^\infty \Big[f(x)-h(x)\Big]^2\,dx=1.88\times 10^{-11}$$
