Asymptotics of $\sum_{i=1}^n \frac{1}{\sqrt{i(i+1)}}$

Question

For sums like $$H_n^r=\sum_{i=1}^n \frac{1}{i^r},$$ from this link we know The Euler-Maclaurin Sum Formula can tackle the asymptotics of such sums, but I find it hard to use the formula to compute even the constant term in the expansion of $$\sum_{i=1}^n \frac{1}{\sqrt{i(i+1)}}.$$ To be more specific, I can't find the regularities of the higher-order derivatives of $\dfrac{1}{\sqrt{x(x+1)}}$, as well as Bernoulli Numbers. Does it take more tricks to calculate these things, or do we have another, better way?

There's no need to look at high derivatives if you're content to write the constant term as the value of an improper integral (as is done for Euler's constant for the traditional harmnoic series) or as the value of an infinite series. (And there's not much reason to think the constant has a simpler closed form.) For example, this constant can be written as $C_0 + D \approx 0.01903$ where $$D = \sum_{k=1}^\infty \biggl( \frac1{\sqrt{k(k+1)}} - \frac1k \biggr) \approx -0.55819.$$ — Greg Martin, Mar 06 '25 at 22:52
You can subtract the harmonic series from it and use a method analogous to this to obtain complete asymptotic expansion. — Gary, Mar 06 '25 at 23:18

score 4 · Accepted Answer · answered Mar 06 '25 at 23:02

You may do like this:

$$ \begin{align*} \sum_{k=1}^{n} \frac{1}{\sqrt{k(k+1)}} &= H_n + \sum_{k=1}^{n} \biggl( \frac{1}{\sqrt{k(k+1)}} - \frac{1}{k} \biggr) \\ &= H_n + D + \sum_{k=n+1}^{\infty} \biggl( \frac{1}{k} - \frac{1}{\sqrt{k(k+1)}} \biggr), \end{align*} $$

where $D = \sum_{k=1}^{\infty} \bigl( \frac{1}{\sqrt{k(k+1)}} - \frac{1}{k} \bigr)$ is the constant mentioned in Greg Martin's comment. Since the sum in the last line is $\mathcal{O}(n^{-1})$, it follows that

$$ \sum_{k=1}^{n} \frac{1}{\sqrt{k(k+1)}} = \log n + \gamma + D + \mathcal{O}(n^{-1}) $$

where $\gamma$ is the Euler–Mascheroni constant. Asymptotic expansions up to an arbitrary order can also be extracted by noting that

$$ \frac{1}{k} - \frac{1}{\sqrt{k(k+1)}} = - \sum_{j=1}^{N} \binom{-1/2}{j} \frac{1}{k^{j+1}} + \mathcal{O}(k^{-N-2}) $$

for any $N \geq 1$.

score 1 · Answer 2 · answered Mar 08 '25 at 08:15

Since we know that $$\int\frac{dk}{\sqrt{k(k+1)}}=2 \tanh ^{-1}\left(\sqrt{\frac{k}{k+1}}\right)$$ we can use Euler-Maclaurin summation formula.

It will give $$\sum_{k=1}^n \frac{1}{\sqrt{k(k+1)}}=\log(n)+(\gamma+D)+\frac{1}{n}-\frac{25}{48 n^2}+\frac{3}{8 n^3}-\frac{2381}{7680n^4}+O\left(\frac{1}{n^5}\right)$$ where $D$ is the constant given by @Greg Martin in comments. If you need more figures $D=-0.558187124483924$. In fact $D$ is given as an infinite summation of weighed zeta functions since $$\frac{1}{\sqrt{k (k+1)}}-\frac{1}{k}=\sum_{m=0}^\infty \frac{a_m}{k^{m+2}}$$ $$\sum_{k=1}^\infty \Bigg(\frac{1}{\sqrt{k (k+1)}}-\frac{1}{k} \Bigg)=\sum_{m=0}^\infty a_m\, \zeta(m+2)$$

score 0 · Answer 3 · answered Mar 09 '25 at 08:47

We have to find the asymptotic behaviour for $n \to \infty$ of the sum

$$s(n) = \sum _{i=1}^{n } \frac{1}{\sqrt{i (i+1)}}$$

Let us adopt a more symmetric approximation to the summand and consider the quantity

$$ds(n):=\sum _{i=1}^{n } \left(\frac{1}{\sqrt{i (i+1)}}-\frac{1}{2} \left(\frac{1}{i}+\frac{1}{i+1}\right)\right)$$

This can be written as

$$ds(n)=s(n) -\frac{1}{2} \left(\left(H_{n+1}-1\right)+H_n\right)$$

where $H_n$ is the harmonic number.

From the asymptotic behaviour of $H(n)$ we obtain

$$s(n) \simeq s_a(n) = c-\frac{1}{2}+\gamma+\frac{1}{n} +\log (n)-\frac{7}{12 n^2}+ \frac{1}{2 n^3} $$

where the constant $c$ is given by

$$c = ds(\infty) \simeq -0.0581871$$

The plot of $s(n)$ and $s_a(n)$ shows surprisingly good aggreement even down to $n=2$

Asymptotics of $\sum_{i=1}^n \frac{1}{\sqrt{i(i+1)}}$

3 Answers3