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Applying the Copson's inequality, I found: $$S=\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2\lt\dfrac{2}{3}\pi^2$$ where $\Psi^{(1)}(k)$ is the polygamma function. Is it known any sharper bound for the sum $S$? Thanks.

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  • $\begingroup$ You can try this bound $\zeta(2)+2\zeta(3)$. $\endgroup$ Jul 30, 2014 at 16:25
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ Mar 12, 2018 at 16:56

4 Answers 4

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First, we have $$ \begin{align} \psi'(n) &=\sum_{k=0}^\infty\frac1{(k+n)^2}\\ &=\sum_{k=n}^\infty\frac1{k^2}\tag{1} \end{align} $$ Then $$ \begin{align} \sum_{n=1}^\infty\psi'(n)^2 &=\sum_{n=1}^\infty\sum_{j=n}^\infty\frac1{j^2}\sum_{k=n}^\infty\frac1{k^2}\tag{2}\\ &=\sum_{n=1}^\infty\left(\sum_{j=n}^\infty\frac1{j^4}+2\sum_{j=n}^\infty\sum_{m=1}^\infty\frac1{j^2}\frac1{(j+m)^2}\right)\tag{3}\\ &=\sum_{j=1}^\infty\sum_{n=1}^j\frac1{j^4}+2\sum_{j=1}^\infty\sum_{m=1}^\infty\sum_{n=1}^j\frac1{j^2}\frac1{(j+m)^2}\tag{4}\\ &=\sum_{j=1}^\infty\frac1{j^3}+2\sum_{j=1}^\infty\sum_{m=1}^\infty\frac1{j(j+m)^2}\tag{5}\\ &=\zeta(3)+2\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}\tag{6}\\ &=\zeta(3)-2\zeta(3)+2\sum_{n=1}^\infty\frac{H_n}{n^2}\tag{7}\\[9pt] &=\zeta(3)-2\zeta(3)+4\zeta(3)\tag{8}\\[18pt] &=3\zeta(3)\tag{9} \end{align} $$ Explanation:
$(2)$: use $(1)$
$(3)$: first sum covers $j=k$ the other $j\lt k$ and $j\gt k$
$(4)$: change order of summation
$(5)$: sum in $n$
$(6)$: $n=j+m$ and $j=n-m$ and sum in $m$
$(7)$: $H_{n-1}=H_n-\frac1n$
$(8)$: equation $(14)$ of this answer with $q=2$
$(9)$: add

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    $\begingroup$ Nice answer without integrals! +1 $\endgroup$ Aug 30, 2014 at 7:59
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You may evaluate your series in closed form.

Here are the steps.

Recall the following standard representation for the digamma function: $$ \psi(x) = -\gamma+\int_0^1 \frac{1 - t^{x-1}}{1 - t}{\rm{d}} x, \quad x>0, $$ giving, by differentiation, $$ \psi'(x) = -\int_0^1 \frac{t^{x-1} \ln t}{1 - t}{\rm{d}} x, \quad x>0. $$ One may deduce $$ \left(\psi'(k)\right)^2 = \int_0^1\int_0^1 \frac{(uv)^{k-1} \ln u\ln v}{(1 - u)(1-v)}{\rm{d}}u \:{\rm{d}} v $$ and $$ \sum_{k=1}^\infty\left(\psi'(k)\right)^2 = \int_0^1 \! \!\int_0^1 \frac{\ln u\ln v}{(1-uv)(1 - u)(1-v)}{\rm{d}} u\:{\rm{d}} v $$ By partial fraction decomposition and successive integrations involving $\text{Li}_{2}(\cdot)$, you arrive at $$ \sum_{k=1}^\infty\left(\psi'(k)\right)^2 = 3\zeta(3) = 3.60617070947878285619921 \cdots. $$

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    $\begingroup$ Very nice: equality is the sharpest bound of all! $\endgroup$ Jul 31, 2014 at 14:57
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    $\begingroup$ @Semiclassical It's a great statement !!!. $\endgroup$ Aug 3, 2014 at 18:22
  • $\begingroup$ Please correct me if I am wrong, but shouldn't the first line be $\psi(x)=-\gamma+\int^1_0\frac{1-t^{x-1}}{1-t}{\rm d}x$ instead? Of course, this doesn't affect the accuracy of your answer in any way. :) $\endgroup$ Aug 31, 2014 at 12:17
  • $\begingroup$ @SuperAbound Typo corrected. Thank you! $\endgroup$ Aug 31, 2014 at 12:30
  • $\begingroup$ This is quite beautiful. $\endgroup$ Sep 11, 2020 at 7:52
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The upper bound can be improved using asymptofic series :

enter image description here

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Continuing from Olivier Oloa's answer,

$$ \begin{align} \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} &= \int_{0}^{1} \int_{0}^{1} \frac{\ln u \ln v}{(1-uv)(1-u)(1-v)} \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} \int_{0}^{1} \left(\frac{\ln u}{1-u} - \frac{v \ln u}{1-vu} \right) \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} \left( \int_{0}^{1} \frac{\ln u}{1-u} \ du - v \int_{0}^{1} \frac{\ln u}{1-vu} \ du\right) \ dv \end{align}$$

where $$ \int_{0}^{1} \frac{\ln u}{1-u} \ du = \int_{0}^{1} \frac{\ln (1-w)}{w} = -\text{Li}_{2}(1) = -\zeta(2) $$

and

$$ \int_{0}^{1} \frac{\ln u}{1-vu} \ du = - \frac{1}{v} \ln(1-vu) \ln u \Bigg|^{1}_{0} + \frac{1}{v} \int_{0}^{1} \frac{\ln (1-vu)}{u} \ du = - \frac{\text{Li}_{2}(v)}{v} .$$

Therefore,

$$ \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} = \int_{0}^{1} \frac{\ln v}{(1-v)^2} \Big(\text{Li}_{2}(v) - \zeta(2) \Big) \ dv .$$

Then integrating by parts,

$$ \begin{align} &= \big(\text{Li}_{2}(v) - \zeta(2) \big) \left(\ln (1-v) + \frac{v \ln v}{1-v} \right)\Bigg|^{1}_{0} + \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dv + \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &= \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dv + \int_{0}^{1} \frac{ \ln(1-v) \ln v}{1-v} \ dv \end{align}$$

where $$ \begin{align} \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dt &= \ln^{2}(1-v)\ln v \Bigg|^{1}_{0} + 2 \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &= 2 \int_{0}^{1} \frac{ \ln(1-v) \ln v}{1-v} \ dv . \end{align} $$

Therefore, $$ \begin{align} \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} &= 3 \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &= -3 \int_{0}^{1} \ln v \sum_{n=1}^{\infty} H_{n}v^{n} \ dv \\ & = -3 \sum_{n=1}^{\infty} H_{n} \int_{0}^{1} v^{n} \ln v \ dv \\ &= 3 \sum_{n=1}^{\infty} \frac{H_{n}}{(n+1)^{2}} \\ &= 3 \left(\sum_{n=1}^{\infty} \frac{H_{n+1}}{(n+1)^{2}} - \sum_{n=1}^{\infty} \frac{1}{(n+1)^{3}} \right) \\ &= 3 \left( \sum_{n=1}^{\infty} \frac{H_{n}}{n^{2}} -1 - \zeta(3) + 1 \right) \\ &= 3 \big(2 \zeta(3) - \zeta(3) \big) \tag{1} \\ &= 3 \zeta(3) .\end{align} $$

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$(1)$ Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$

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  • $\begingroup$ Thank you for these details! $\endgroup$ Aug 3, 2014 at 17:31
  • $\begingroup$ @OlivierOloa You're welcome. $\endgroup$ Aug 3, 2014 at 17:37

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