<s>[INST] 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).
 [/INST] (This is not a solution.)
The question of convergence is raised in the comments, so I show a proof here.
From $k > 0$ and
$$  k \leq j \leq k^2 $$
we have
$$ \frac{1}{\sqrt{k}} \geq \frac{1}{\sqrt{j}} \geq \frac{1}{k}  \text{,}  $$
using the monotonicity of the square root and standard results about reciprocals and inequalities.  So $1/\sqrt{j}$ is lower bounded by $1/k$.  Then
\begin{align*}
\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}  &\geq  \sum_{j=k}^{k^2} \frac{1}{k} = (k^2 - k + 1)\frac{1}{k}  \text{,}  \\
    \left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2  &\geq \frac{(k^2 - k + 1)^2}{k^2}  \text{,}  \\
    \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2}  &\leq \frac{k^2}{(k^2 - k + 1)^2}  \text{, and}  \\
    \sum_{k=1}^\infty \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2}  &\leq \sum_{k=1}^\infty\frac{k^2}{(k^2 - k + 1)^2}  \text{.}
\end{align*}
(This last sum can be expressed in terms of polygamma functions, but let's not.)
Factoring out "the big part", we have
$$  \frac{k^2}{(k^2 - k + 1)^2} = \frac{k^2}{k^4} \cdot \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4}  \text{.}  $$
This is $\frac{1}{k^2} \cdot \text{[some bounded expression]}$.  Let $f(k_) = \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4}$.  For $k \geq 1$, $f'(k) = 0$ at $k = 2$ and $f''(2) > 0$.  So $f'(k) < 0$ on $[1,2)$ and $f'(k) > 0$ on $(2,\infty)$.  This means the value of $f(k)$ on $[1,\infty)$ is bounded by
$$
\max \{f(1), \lim_{k \rightarrow \infty} f(k)\} = \max \{1,1\} = 1  \text{.}  $$
So \begin{align*}
&\sum_{k=1}^\infty \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2}  \\
&\leq \sum_{k=1}^\infty \frac{1}{k^2} \cdot \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4}  \\
&\leq \sum_{k=1}^\infty \frac{1}{k^2} \cdot 1  \\
&= \frac{\pi^2}{6}  \text{.}  
\end{align*}
(This last is the Basel problem.  Alternatively, one can also use the comparison test with $\int^\infty \frac{\mathrm{d}x}{x^2}$ to show this last sum converges.)
So the sum converges.  There's a bit of a gap between this upper bound and the partial sum you cite, but our first approximation (replacing all the $1/\sqrt{j}$ terms with the extremal term in their sum) is a loose step and pretending the bounded constant near the end is $1$ is also loose.
</s>