Nested sums - reciprocal of a sum : Find exact value

Question

$$\text{Find:} ~~~~~~ \sum_{k=1}^{\infty} \frac{1}{ \left ( \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right )^2}$$

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(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).

Answer 1

(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.

Answer 2

As suggested in my comment, to prove convergence (this approach can only however help you approximate the value of the sum, not get it exactly — I doubt there is a nice closed form), you can write $$ \int_{j}^{j+1} \frac{dx}{\sqrt{x}} \leq \int_{j}^{j+1} \frac{dx}{\sqrt{j}} = \frac{1}{\sqrt{j}} = \int_{j-1}^{j} \frac{dx}{\sqrt{j}} \leq \int_{j-1}^{j} \frac{dx}{\sqrt{x}} $$ and so, for $k\geq 2$, $$ 2k-2\sqrt{k}\leq \int_{k}^{k^2+1} \frac{dx}{\sqrt{x}} = \sum_{j=k}^{k^2} \int_{j}^{j+1} \frac{dx}{\sqrt{x}} \leq \sum_{j=k}^{k^2} \frac{1}{\sqrt{j}} \leq \sum_{j=k}^{k^2} \int_{j-1}^{j} \frac{dx}{\sqrt{x}} = \int_{k-1}^{k^2} \frac{dx}{\sqrt{x}} \leq 2k \tag{1} $$ using that $\int_{a}^{b} \frac{dx}{\sqrt{x}} = 2(\sqrt{b}-\sqrt{a})$. Since $k-\sqrt{k}\geq k/2$ for $k\geq 4$, we have $$ \sum_{j=k}^{k^2} \frac{1}{\sqrt{j}} \geq k $$ for $k\geq 4$, and therefore $$ \sum_{k=4}^\infty \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2} \leq \sum_{k=4}^\infty \frac{1}{k^2} < \infty \tag{2} $$ which shows convergence.

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Note that using (1), you could approximate the original sum arbitrarily by using the inequalities from (1) starting at any given $k=K$ of your choosing, and computing exactly the value of the terms for $k\leq K$. That does not seem very practical, however.

Answer 3

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 $$a_k=2 k-2 \sqrt{k}+\frac 1{2\sqrt k}+\frac 1{2k}+\frac 1{24 k\sqrt k}-\frac 1{24k^3}+O\left(\frac{1}{k^{7/2}}\right)$$ $$\frac{1}{ a_k^2}=\frac 1{4k^2}+\frac 1{2k^{5/2}}+\frac 3{4k^{3}}+O\left(\frac{1}{k^{7/2}}\right)$$ $$\sum_{k=1}^{\infty} \frac{1}{ a_k^2}=\sum_{k=1}^{p-1} \frac{1}{ a_k^2}+\sum_{k=p}^{\infty} \frac{1}{ a_k^2}$$

Answer 4

I thought I might expand on my comment from a while ago regarding the value of the sum.

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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.

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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 $$ \sum_{k=1}^\infty\frac1{\left(\sum\limits_{j=k}^{k^2}\frac1{\sqrt{j\,}}\right)^2}=1.596346969939244669227454284962\tag8 $$