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Is there a general formula for the following

$$\sum_{n\geq 1}\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\cdots \pm \frac{1}{n}\right)}{n^p}\,\, p\geq 1$$

What about some restrictions on $p$ , like integers or anything helpful ?

Zaid Alyafeai
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  • Why not simply let $p>0$? – Eric Auld Aug 05 '13 at 08:19
  • @EricAuld You can prove that this forluma$\ge\sum \frac{1}{2n^p}$. – asatzhh Aug 05 '13 at 09:05
  • @asatzhh Whoops! I didn't see Zaid wanted the infinite sum! In that case, it should be $p>1$, right? – Eric Auld Aug 05 '13 at 13:59
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    @EricAuld Yes, it should be. And I guess there is no closed formula for this. – asatzhh Aug 05 '13 at 14:32
  • Has this been done elsewhere? :) If $p$ is odd positive integer I suppose there is a simple closed form, however for even $p$ I'm getting stuck with sth like $\displaystyle \int_0^{-1} \frac{\operatorname{Li}_{s}^2(x)}{x},dx$ where $p = 2s$. Is a closed form for the later integral known/manageable? :) – r9m May 16 '15 at 14:06

1 Answers1

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There are closed forms for any integer $p \ge 2$.(Note that for $p=1$ this sum does not converge because the sum in the numerator of the summand tends for big $n$ to $\log(2)$ and thus we end up with a harmonic series which diverges).

We denote the quantity in question as follows: \begin{eqnarray} {\tilde {\bf H}}^{(1)}_p(+1) &:=& \sum\limits_{m=1}^\infty \frac{{\tilde H}^{(1)}_m}{m^p} \cdot (+1)^m \\ { {\bf H}}^{(1)}_p(+1) &:=& \sum\limits_{m=1}^\infty \frac{{ H}^{(1)}_m}{m^p} \cdot (+1)^m \end{eqnarray} where ${\tilde H}^{(q)}_n:=\sum\limits_{m=1}^n (-1)^{m-1}/m^q$.

By order the order of summation in the definition above we immediately get: \begin{eqnarray} {\tilde {\bf H}}^{(1)}_p(+1) = \zeta(p) \cdot \log(2)-Li_{p+1}(-1) + {\bf H}^{(p)}_1(-1) \end{eqnarray} However the quantity in bold font in the very right hand side is perfectly known. We have: \begin{eqnarray} {\bf H}^{(2p+1)}_1(-1) &=& Li_{2p+2}(-1) + \sum\limits_{j=1}^p Li_j(-1) Li_{2p+2-j}(-1) (-1)^{j-1} + \frac{1}{2} [Li_{p+1}(-1)]^2 \\ {\bf H}^{(2p+0)}_1(-1) &=& \left\{ \begin{array}{ll} -\zeta(3)+ \frac{1}{2} \log(2) \zeta(2) & \mbox{if $p=1$}\\ -2 \zeta(5) + \frac{3}{8} \zeta(2) \zeta(3) + \frac{7}{8} \log(2) \zeta(4) & \mbox{if $p=2$}\\ -3 \zeta(7)+\frac{15}{32} \zeta(2) \zeta(5) + \frac{21}{32} \zeta(4) \zeta(3)+\frac{31}{32} \log(2) \zeta(6) & \mbox{if $p=3$}\\ -4 \zeta(9) + \frac{63}{128} \zeta(2) \zeta(7)+\frac{105}{128} \zeta(4) \zeta(5) + \frac{93}{128}\zeta(6) \zeta(3)+\frac{127}{128} \log(2) \zeta(8) & \mbox{if $p=4$} \end{array} \right. \end{eqnarray} where the last line follows from Calculating alternating Euler sums of odd powers . Bringing everything together we get: \begin{eqnarray} {\tilde {\bf H}}^{(1)}_p(+1) &=& \left\{ \begin{array}{rr} \frac{1}{4} (\zeta(2) \log (64)-\zeta (3)) & \mbox{if $p=2$}\\ \frac{7}{4} \zeta (3) \log (2)-\frac{5 \zeta(4)}{16} & \mbox{if $p=3$}\\ \frac{1}{16} (6 \zeta(2) \zeta (3)+30 \zeta(4) \log (2)-17 \zeta (5)) & \mbox{if $p=4$}\\ \frac{1}{64} \left(-49 \zeta(6)+18 \zeta (3)^2+124 \zeta (5) \log (2)\right) & \mbox{if $p=5$}\\ -\frac{3}{64} (-10 \zeta(2) \zeta (5)-14 \zeta(4) \zeta (3)-42 \zeta(6) \log (2)+43 \zeta (7)) & \mbox{if $p=6$}\\ \frac{1}{256} (-321 \zeta(8)+180 \zeta (3) \zeta (5)+508 \zeta (7) \log (2)) & \mbox{if $p=7$}\\ \frac{1}{256} (126 \zeta(2) \zeta (7)+210 \zeta(4) \zeta (5)+186 \zeta(6) \zeta (3)+510 \zeta(8) \log (2)-769 \zeta (9)) & \mbox{if $p=8$}\\ \frac{-1793 \zeta(10)+756 \zeta (3) \zeta (7)+450 \zeta (5)^2+2044 \zeta (9) \log (2)}{1024} & \mbox{if $p=9$} \end{array} \right. \end{eqnarray}

Przemo
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