2
$\begingroup$

Generalizing a recent post Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ where convergence was assured by alternating the sign here's a similar problem in which convergence in forced by the increased power in the denominator.

Question: is there a closed form of this sum

$$\begin{align} s_2 &=\sum_{k=1}^{\infty} \frac{\left\lfloor \sqrt{k}\right\rfloor}{k^2}\simeq 2.33198\tag{1}\\ \end{align}\tag{1}$$

The sum is obviously convergent, and obeys the following inequality

$$1.64493\simeq\zeta(2)=\sum_{k=1}^{\infty} \frac{1}{k^2}\lt s_{2} \lt \sum_{k=1}^{\infty} \frac{\sqrt{k}}{k^2}=\zeta(\frac{3}{2})\simeq 2.61238\tag{2}$$

$\endgroup$
3
$\begingroup$

My solution attempt

I have not found a closed form expression but the following integral representation

$$s_{2} =\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right \rfloor}{k^2} = \int_0^{\infty } \frac{t \left(\vartheta _3\left(0,e^{-t}\right)-1\right)}{2 \left(1-e^{-t}\right)} \, dt\tag{1}$$

Here

$$\vartheta _3(u,q)=1+2 \sum _{n=1}^{\infty } q^{n^2} \cos (2 n u)$$

is a Jacobi theta function.

Derivation

The drivation starts with a similar method as in $[1]$.

We find that the partial sum from $k=1$ to $k=m^2-1$ ($m \in N$) can be written as

$$\begin{align} p(m) &= \sum_{k=1}^{m^2-1} \frac{\left\lfloor \sqrt{k}\right\rfloor}{k^2}= f(m) - g(m) \end{align}\tag{2}$$

where

$$\begin{align} f(m) & =m H(m^2-1,2)\tag{3}\\ g(m) &= \sum_{k=1}^{m} H(k^2-1,2)\tag{4} \end{align}$$

Here $H(n,2)=H_{n,2}=\sum_{k=1}^n \frac{1}{k^2}$ is the generalized harmonic number of order $2$ of $n$.

Indeed, writing (dropping the second index $2$ in $H$ for simplicity)

$m=2\to k=1..3$ :
$\frac{\left\lfloor \sqrt{1} \right \rfloor}{1^2}+\frac{\left\lfloor \sqrt{2} \right \rfloor}{2^2}+\frac{\left\lfloor \sqrt{3} \right \rfloor}{3^2}= \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}= H_{3}= H(2^2-1)$

$m=3\to k=4..8$ : $\frac{\left\lfloor \sqrt{4} \right \rfloor}{4^2}+\frac{\left\lfloor \sqrt{5} \right \rfloor}{5^2}+\frac{\left\lfloor \sqrt{6} \right \rfloor}{6^2}+\frac{\left\lfloor \sqrt{7} \right \rfloor}{7^2}+\frac{\left\lfloor \sqrt{8} \right \rfloor}{8^2}=2 \frac{1}{4^2}+2\frac{1}{5^2}+2\frac{1}{6^2}+2\frac{1}{7^2}+2\frac{1}{8^2}= 2(H(8)-H(3))=2 (H(3^2-1)-H(2^2-1))$

Together up to $m=3$

$p(3) = H(3) + 2 (H(8)-H(3))=2 H(8) - H(3) $

and so on gives an the first place

$p(m) = (m-1) H(m^2-1) - \sum_{k=1}^{m-1} H(k^2-1)$

but then shifting $-H(m^2-1)$ from the first term to the second, i.e. including it into the sum we get $(2)$,$(3)$ and $(4)$.

Now we need the limit $m\to\infty$.

This is no problem for $f$ where we have

$$f(m) \sim \zeta(2)-\frac{1}{m}-\frac{1}{2 m^3}+ O(\frac{1}{m^5})\tag{5}$$

Now since

$$ H(k^2-1,2) = \sum_{j=1}^{k^2-1} \frac{1}{j^2} = \sum_{j=1}^{\infty} \frac{1}{j^2}-\sum_{j=k^2}^{\infty} \frac{1}{j^2}=\zeta(2)-\sum_{j=k^2}^{\infty} \frac{1}{j^2} $$

$g$ can be written as

$$g(m) = m \zeta(2) - \sum_{k=1}^m \sum_{j=k^2}^\infty \frac{1}{j^2}\tag{6}$$

Hence we have

$$s_{2}=\lim_{m\to \infty } \, p(m) =\lim_{m\to \infty } \,(\sum_{k=1}^m \sum_{j=k^2}^\infty \frac{1}{j^2})\tag{7} $$

and we have to calculate the asymptotic behaviour of the double sum

$$g_1(m) = \sum_{k=1}^m \sum_{j=k^2}^\infty \frac{1}{j^2}\tag{8}$$

With

$$\frac{1}{j^2}=\int_0^{\infty } t \exp (-j t) \, dt$$

we can do the $j$-sum

$$\sum _{j=k^2}^{\infty } \exp (-j t)=\frac{e^{-k^2 t}}{1-e^{-t}}$$

and subsequently do the $k$-sum extending the limit $m\to\infty$

$$\sum _{k=1}^{\infty } e^{-k^2 t}=\frac{1}{2} \left(\vartheta _3\left(0,e^{-t}\right)-1\right)$$

Putting this back into the the $t$-integral gives $(1)$ QED.

Discussion

In the previous problem $[1]$ other users have provided interesting results with other approaches to "remove" the floor function. I'm sure this can be done here as well.

Maybe also a head-on attack on the double sum $(8)$ can lead to "sum-based" simplifications.

Proximity to the harmonic numbers makes the existence of a closed form for $s_2$ probable.

References
$[1]$ Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$

$\endgroup$
2
  • $\begingroup$ How do you have (2), (3), and (4)? I don't see how they are true... $\endgroup$ – clathratus Nov 29 '19 at 15:50
  • 2
    $\begingroup$ @ clathratus Sorry, I thought to save space by referencing the solution of the previous problem $[1]$. But I will elaborate soon (I'm in a hurry at the moment). $\endgroup$ – Dr. Wolfgang Hintze Nov 29 '19 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.