Closed expression for sum $\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right \rfloor}{k^2}$ 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
$$1.64493\simeq\zeta(2)=\sum_{k=1}^{\infty} \frac{1}{k^2}\lt s_{2} \lt \sum_{k=1}^{\infty} \frac{\sqrt{k}}{k^2}=\zeta(\frac{3}{2})\simeq 2.61238\tag{2}$$
 A: 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 $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$
