Nested sums - reciprocal of a sum : Find exact value $$\text{Find:} ~~~~~~ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2}$$

(Beware of the bounds of $j$, it does not always start from $1$, but starts at $k$ and goes to $k^2$)
At first I thought it has something to do with Riemann's zeta function $\zeta$ and thus the solution is:
$$ \zeta \left( \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right ) $$
However this totally seems incorrect because we cannot find this sum as we have an unknown, $k$.
Which also is being summed to $\infty$.
I thought (however was unsure) of the fact that:
$$ \left ( \sum f \right )^2 \ge  \sum f^2  $$
Thus (maybe) it is trivial the sum is converging to some value, which according to a little program I wrote is about: $$ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2} \approx 1.596$$
But this is not a rigorous proof, is there a way to solve this so we get a solution by hand - calculating the exact value of this sum if it is finite (if not, why? )
Any mathematical tools (not programs) are acceptable, because this is not a question from a specific course ( I don't have context for this).

Let
$$a_k=\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}=H_{k^2}^{\left(\frac{1}{2}\right)}-H_{k-1}^{\left(\frac{1}{2}\right)}  $$ and
$$S_p=\sum_{k=1}^{p} \frac{1}{ a_k^2}$$
Using asymptotics of the harmonic numbers and continuing with Taylor series
$$a_k=2 k-2 \sqrt{k}+\frac 1{2\sqrt k}+\frac 1{2k}+\frac 1{24 k\sqrt k}-\frac 1{24k^3}+O\left(\frac{1}{k^{7/2}}\right)$$
$$\frac{1}{ a_k^2}=\frac 1{4k^2}+\frac 1{2k^{5/2}}+\frac 3{4k^{3}}+O\left(\frac{1}{k^{7/2}}\right)$$
$$\sum_{k=1}^{\infty} \frac{1}{ a_k^2}=\sum_{k=1}^{p-1} \frac{1}{ a_k^2}+\sum_{k=p}^{\infty} \frac{1}{ a_k^2}$$
