28
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You can try this bound $\zeta(2)+2\zeta(3)$. $\endgroup$ – Mhenni Benghorbal Jul 30 '14 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. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 16:56
21
$\begingroup$

The upper bound can be improved using asymptofic series :

enter image description here

$\endgroup$
31
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Nice answer without integrals! +1 $\endgroup$ – Olivier Oloa Aug 30 '14 at 7:59
24
$\begingroup$

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 $$ \sum_{k=1}^\infty\left(\psi'(k)\right)^2 = 3\zeta(3) = 3.60617070947878285619921 \cdots. $$

$\endgroup$
  • 4
    $\begingroup$ Very nice: equality is the sharpest bound of all! $\endgroup$ – Semiclassical Jul 31 '14 at 14:57
  • 2
    $\begingroup$ @Semiclassical It's a great statement !!!. $\endgroup$ – Felix Marin Aug 3 '14 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. :) $\endgroup$ – SuperAbound Aug 31 '14 at 12:17
  • $\begingroup$ @SuperAbound Typo corrected. Thank you! $\endgroup$ – Olivier Oloa Aug 31 '14 at 12:30
20
$\begingroup$

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

$ $

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

$\endgroup$
  • $\begingroup$ Thank you for these details! $\endgroup$ – Olivier Oloa Aug 3 '14 at 17:31
  • $\begingroup$ @OlivierOloa You're welcome. $\endgroup$ – Random Variable Aug 3 '14 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.