TITLE: Nested sums - reciprocal of a sum : Find exact value
QUESTION [8 upvotes]: $$\text{Find:} ~~~~~~ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2}$$

(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).

REPLY [2 votes]: I thought I might expand on my comment from a while ago regarding the value of the sum.

Convergence
Considering the number of terms times the smallest term, we get
$$
\begin{align}
\sum_{j=k}^{k^2}\frac1{\sqrt{j\,}}
&\ge\left(k^2-k+1\right)\frac1k\tag{1a}\\
&\ge k-1\tag{1b}
\end{align}
$$
Therefore, since the term for $k=1$ is $1$,
$$
\sum_{k=1}^\infty\frac1{\left(\sum\limits_{j=k}^{k^2}\frac1{\sqrt{j\,}}\right)^2}
\le1+\sum_{k=2}^\infty\frac1{(k-1)^2}\tag2
$$
Since the sum on the right-hand side of $(2)$ converges, the sum on the left-hand side of $(2)$ also converges.

Value
The Euler-Maclaurin Sum Formula says
$$
\begin{align}
\sum_{j=1}^k\frac1{\sqrt{j\,}}
&=\zeta\!\left(\frac12\right)+2\sqrt{k}+\frac1{2\sqrt{k}}-\frac1{24k^{3/2}}+\frac1{384k^{7/2}}-\frac1{1024k^{11/2}}\\
&{}+\frac{143}{162840k^{15/2}}-\frac{1105}{786432k^{19/2}}+O\!\left(\frac1{k^{23/2}}\right)\tag3
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{j=1}^{k^2}\frac1{\sqrt{j\,}}
&=\zeta\!\left(\frac12\right)+2k+\frac1{2k}-\frac1{24k^3}+\frac1{384k^7}-\frac1{1024k^{11}}\\
&{}+\frac{143}{162840k^{15}}-\frac{1105}{786432k^{19}}+O\!\left(\frac1{k^{23}}\right)\tag4
\end{align}
$$
Thus, subtracting $(3)$ from $(4)$ and adding $\frac1{\sqrt{k}}$ back in,
$$
\begin{align}
\scriptsize\sum_{j=k}^{k^2}\frac1{\sqrt{j\,}}
&\scriptsize=2k-2\sqrt{k}+\frac1{2\sqrt{k}}+\frac1{2k}+\frac1{24k^{3/2}}-\frac1{24k^3}-\frac1{384k^{7/2}}+\frac1{1024k^{11/2}}\\
&\scriptsize{}+\frac1{384k^7}-\frac{143}{162840k^{15/2}}+\frac{1105}{786432k^{19/2}}-\frac1{1024k^{11}}+O\!\left(\frac1{k^{23/2}}\right)\tag5
\end{align}
$$
Squaring and taking the reciprocal of $(5)$ yields
$$
\begin{align}
\scriptsize\frac1{\left(\sum\limits_{j=k}^{k^2}\frac1{\sqrt{j\,}}\right)^2}
&\scriptsize=\frac1{4k^2}+\frac1{2k^{5/2}}+\frac3{4k^3}+\frac7{8k^{7/2}}+\frac3{4k^4}+\frac{35}{96k^{9/2}}-\frac{15}{64k^5}-\frac{29}{32k^{11/2}}-\frac{185}{128k^6}\\
&\scriptsize{}-\frac{2555}{1536k^{13/2}}-\frac{4409}{3072k^7}-\frac{581}{768k^{15/2}}+\frac{485}{2048k^8}+\frac{15923}{12288k^{17/2}}+\frac{17359}{8192k^9}+\frac{67421}{27648k^{19/2}}\\[6pt]
&\scriptsize{}+\frac{930967}{442368k^{10}}+\frac{6741487}{5898240k^{21/2}}-\frac{2771203}{11796480k^{11}}-\frac{14834047}{8847360k^{23/2}}-\frac{590040679}{212336640k^{12}}\\[6pt]
&\scriptsize{}-\frac{90602227}{28311552k^{25/2}}-\frac{260952131}{94371840k^{13}}-\frac{16220531}{10616832k^{27/2}}+\frac{384274163}{1698693120 n^{14}}+O\!\left(\frac1{k^{29/2}}\right)\tag6
\end{align}
$$
Apply the Euler-Maclaurin Sum Formula to $(6)$:
$$
\begin{align}
\scriptsize\sum_{k=1}^n\frac1{\left(\sum\limits_{j=k}^{k^2}\frac1{\sqrt{j\,}}\right)^2}
&\scriptsize=C-\frac1{4n}-\frac1{3n^{3/2}}-\frac1{4n^2}-\frac1{10n^{5/2}}+\frac1{12n^3}+\frac{11}{48n^{7/2}}+\frac{63}{256n^4}+\frac{37}{288n^{9/2}}\\
&\scriptsize{}-\frac{67}{960n^5}-\frac{2197}{8448n^{11/2}}-\frac{5959}{18432n^6}-\frac{6479}{33280 n^{13/2}}+\frac{647}{7168n^7}+\frac{12539}{30720n^{15/2}}\\[6pt]
&\scriptsize{}+\frac{330233}{589824n^8}+\frac{703451}{1880064n^{17/2}}-\frac{1642441}{9953280n^9}-\frac{145034701}{168099840n^{19/2}}-\frac{455987831}{353894400n^{10}}\\[6pt]
&\scriptsize{}-\frac{116451047}{123863040n^{21/2}}+\frac{490021111}{1167851520n^{11}}+\frac{266366981}{108527616n^{23/2}}+\frac{13475716393}{3397386240n^{12}}\\[6pt]
&\scriptsize{}+\frac{6651802811}{2123366400n^{25/2}}-\frac{113110493771}{77290536960n^{13}}+O\!\left(\frac1{n^{27/2}}\right)\tag{7}
\end{align}
$$
We then compute $C$ by evaluating the left-hand side of $(7)$ for a large $n$, then comparing with the right-hand side.
Using $n=300$, we get
$$
\sum_{k=1}^\infty\frac1{\left(\sum\limits_{j=k}^{k^2}\frac1{\sqrt{j\,}}\right)^2}=1.596346969939244669227454284962\tag8
$$