[Edit: rewritten 4/8/22 to increase specificity of question]

Recently, I was interested to estimate the following partial sum.


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 $\frac{1}{\sqrt{x(x+1)}} \sim \frac 1x$?

  • 2
    $\begingroup$ Because $\lim_{x\to\infty }\sqrt{x(x+1)}\ln\left(\frac{x+1}{x}\right)=1$. $\endgroup$
    – Surb
    Dec 3, 2021 at 17:10
  • 5
    $\begingroup$ Your argument IS the reason why the functions coincide so well for large $x$. I don't get why you are not content with this argument. $\endgroup$
    – Peter
    Dec 3, 2021 at 17:20
  • 3
    $\begingroup$ It's not really possible to answer this question without more information about what kind of explanation the OP is seeking. $\endgroup$ Dec 3, 2021 at 17:23
  • 2
    $\begingroup$ It's not very much of a coincidence when very natural functions have Taylor coefficients which are simple fractions. It's the Law of Small Numbers. $\endgroup$ Dec 3, 2021 at 17:48

3 Answers 3


I don't think there is anything special about the fact that the function $f(x) = ln(1 + 1/x)$ is -for large $x$- so well approximated by your expression. $f(x)$ is well-behaved and monotically decreassing. Its Taylor series for large $x$ can be derived. It is then possible to construct other functions $g(x)$ that have the same first few terms of the Taylor series. For example, I expect this function to work even better:

$$g(x) = \frac{x+1/6}{x(x+2/3)}$$

The reason is that the error between $f$ and $g$ for large $x$ is now of order $x^{-4}$.


Note that for $x> 0$ we have

$$\sqrt{x (x+1)} \ln\left( \frac{x+1}{x} \right) = \ln\left(\left(\frac{x+1}{x} \right)^{\sqrt{x (x+1)}}\right) = \ln\left(\left(1 + \frac{1}{x} \right)^{\sqrt{x (x+1)}}\right).$$

Now, as the $x^2$ dominates, we have $x \sim \sqrt{x (x+1)}$, so

$$\ln\left(\left(1 + \frac{1}{x} \right)^{\sqrt{x (x+1)}}\right) \sim \ln\left(\left(1 + \frac{1}{x} \right)^{x}\right).$$

(This step could be made more formal, but I will skip that here.)

And we're nearly done: Using the continuity of $\ln$, we finally get $$\sqrt{x (x+1)} \ln\left( \frac{x+1}{x} \right) \sim \ln\left(\left(1 + \frac{1}{x} \right)^{x}\right) \to \ln(\text{e}) = 1, x \to \infty.$$

I hope this is somewhat satisfying.


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 $$\Phi_1=\int_{10}^\infty \Big[f(x)-g(x)\Big]^2\,dx=2.73\times 10^{-9}$$ $$\Phi_2=\int_{10}^\infty \Big[f(x)-h(x)\Big]^2\,dx=1.88\times 10^{-11}$$


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