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

(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).

  • $\begingroup$ Based on the end of your question, it is not clear to me whether you want to compute the sum exactly (might be hard) or show it is finite (much easier). $\endgroup$
    – Clement C.
    Feb 12, 2021 at 23:10
  • $\begingroup$ @ClementC. I will edit the question so it would be clearer, I want to compute the exact value by hand if it is finite (otherwise there is no point, but, I couldn't show it is diverging yet) $\endgroup$
    – CSch of x
    Feb 12, 2021 at 23:11
  • 2
    $\begingroup$ To show it is finite, do a comparison with an integral on the sum in the denominator to show it is of order $2\sqrt{k^2}-2\sqrt k \approx 2k$. Then, the thing is squared, so you can do a comparison with $1/k^2$. $\endgroup$
    – Clement C.
    Feb 12, 2021 at 23:13
  • $\begingroup$ Numerically, using the Euler-Maclaurin Sum Formula twice, I get $1.596346969939244669227454284962$. $\endgroup$
    – robjohn
    Feb 13, 2021 at 5:31

4 Answers 4


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


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.


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


(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.


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.

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.


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

  • $\begingroup$ (+1) $(5)$ and $(6)$ from my answer match $a_k$ and $a_k^{-2}$ above. I am not sure if what you were suggesting above is what I have posted. $\endgroup$
    – robjohn
    Apr 12, 2021 at 17:53
  • $\begingroup$ Just out of curiosity, how did you get the asymptotic expansion for $a_k$? $\endgroup$
    – robjohn
    Apr 12, 2021 at 21:10
  • $\begingroup$ @robjohn. I used the asymptotics of $H_{p}^{\left(\frac{1}{2}\right)}$, used it twice for $p=k^2$ and $p=k-1$, replace and continue with Taylor. This gives $a_k$. This is close to what you did. Cheers :-) $\endgroup$ Apr 13, 2021 at 2:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .