Is the closed form for $\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}$ known in the literature?

I managed to find

$$\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=(1-2^{-m})\sum_{k=1}^\infty\frac{H_k}{k^m}-2^{-m}\sum_{k=1}^\infty\frac{H_k}{(k+1/2)^m}$$ $$=(1-2^{-m})\left[\left(1+\frac m2\right)\zeta(m+1)-\frac12\sum_{i=1}^{m-2}\zeta(i+1)\zeta(m-i)\right]$$ $$+\frac{(-2)^{-m-1}}{(m-1)!}\left[2\gamma\ \psi^{(m-1)}\left(\frac12\right)-\psi^{(m)}\left(\frac12\right)+\lim_{\substack{a\to1/2}}\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$

Where $$\overline{H}_k$$ is the skew harmonic number, $$\gamma$$ is Euler–Mascheroni constant, $$\zeta$$ is the Riemann zeta function and $$\psi^{(m)}(a)$$ is the Polylogarithm function where

$$\psi^{(m)}\left(\frac12\right)=(-1)^mm!(1-2^{m+1})\zeta(m+1)$$

My question is the closed form above known in the literature? and can we do further simplifications for the limit term to have a cleaner closed form? Also I would like to see different approaches.

Thank you

Proof

$$\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=1+\sum_{k=2}^\infty\frac{\overline{H}_k}{k^m}=1+\sum_{k=1}^\infty\frac{\overline{H}_{2k}}{(2k)^m}+\sum_{k=1}^\infty\frac{\overline{H}_{2k+1}}{(2k+1)^m}$$

By writing $$\overline{H}_{2k}=H_{2k}-H_k$$ and $$\overline{H}_{2k+1}=H_{2k+1}-H_k$$ we have

$$\sum_{k=1}^\infty\frac{\overline{H}_{2k}}{(2k)^m}=\sum_{k=1}^\infty\frac{H_{2n}}{(2n)^m}-\sum_{n=1}^\infty\frac{H_{n}}{(2n)^m}=\frac12\sum_{k=1}^\infty\frac{(-1)^kH_{k}}{k^m}+\left(\frac12-2^{-m}\right)\sum_{k=1}^\infty\frac{H_{k}}{k^4}$$

and

$$\sum_{k=1}^\infty\frac{\overline{H}_{2k+1}}{(2k+1)^m}=\color{blue}{\sum_{k=1}^\infty\frac{H_{2k+1}}{(2k+1)^m}}-\sum_{k=1}^\infty\frac{H_k}{(2n+1)^m}$$

$$=\color{blue}{-1+\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^m}}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}$$

$$=\color{blue}{-1+\frac12\sum_{k=0}^\infty\frac{(-1)^kH_{k+1}}{(k+1)^m}+\frac12\sum_{k=0}^\infty\frac{H_{k+1}}{(k+1)^m}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^m}$$

$$=\color{blue}{-1-\frac12\sum_{k=1}^\infty\frac{(-1)^kH_{k}}{k^m}+\frac12\sum_{k=1}^\infty\frac{H_{k}}{k^m}}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}\\$$

Combine the two sums,

$$\Longrightarrow \sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=(1-2^{-m})\sum_{k=1}^\infty\frac{H_k}{k^m}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}\tag1$$

The first sum is well-known

$$\sum_{k=1}^\infty\frac{H_k}{k^m}=\left(1+\frac m2\right)\zeta(m+1)-\frac12\sum_{i=1}^{m-2}\zeta(i+1)\zeta(m-i)$$

For the second sum, from here we have

$$\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=(-1)^{m-1}m!\sum_{k=1}^\infty\frac{H_k}{(k+n+1)^{m+1}}\\=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right),\quad n\in\mathbb{R}\ge-1,\quad m\in\mathbb{N}$$

Let $$m+1\to m$$ and $$n+1=a$$ we get

$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}(H_{a-1}^2+H_{a-1}^{(2)})$$

Substitute $$H_{a-1}=\psi(a)+\gamma$$ and $$H_{a-1}^{(2)}=\zeta(2)-\psi^{(1)}(a)$$

$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}((\psi(a)+\gamma)^2+\zeta(2)-\psi^{(1)}(a))$$

Because $$m\ge 2$$ for convergence, we can ignore the constants $$\gamma$$ and $$\zeta(2)$$ on the right side,

$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}(\psi(a)^2-\psi^{(1)}(a)+2\gamma\ \psi(a))$$

$$=\frac12\left[2\gamma\ \psi^{(m-1)}(a)-\psi^{(m)}(a)+\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$

Now take the limit to both sides and let $$a\to 1/2$$ we get

$$\sum_{k=1}^\infty\frac{H_k}{(k+1/2)^m}=\frac{(-1)^m}{2(m-1)!}\left[2\gamma\ \psi^{(m-1)}\left(\frac12\right)-\psi^{(m)}\left(\frac12\right)+\lim_{\substack{a\to1/2}}\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$

By combining the results of the two sums, the closed form follows.

Note

I am tagging " integration" as logarithmic integrals and harmonic series are strongly related.

• You may check Lemma $4$ in Cornel's article here researchgate.net/publication/… to see how to handle with the second series in the right-hand side. The cases with $2m+1$ may be done in a very similar way. Feb 17 '20 at 12:45
• Thank you but the exponent is m not 2m+1. I got your point that we already have the even case and we can similarly get the odd case but I need the generalized m case. Is it possible following the same approach? Feb 17 '20 at 15:46
• Yes, it is possible, but splitting according to $m$ odd and even looks better. Given the formula from the answer below, I think all can be nicely arranged even without splitting according to parity. Feb 17 '20 at 16:07

Yes, a closed form in the literature is known. For $$m \geqslant 2$$ it is: $$\sum_{k = 1}^\infty \frac{\overline H_k}{k^m} = \zeta (m) \log 2 - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1).$$ Here $$\eta (s) = \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^s} = (1 - 2^{1 - s}) \zeta (s)$$ is the Dirichlet eta function and $$\zeta (s) = \sum_{n = 1}^\infty \frac{1}{n^s}$$ is the Riemann zeta function.

References:

1. R. Sitaramachandrarao, "A formula of S. Ramanujan," Journal of Number Theory, 25, 1-19 (1987). See Theorem 3.5 on page 9.

2. Philippe Flajolet and Bruno Salvy, "Euler sums and contour integral representations," Experimental Mathematics, 7(1), 15-35 (1998). See Theorem 7.1 (i) on page 32.

• Thank you this is much simpler. +1 Feb 17 '20 at 3:51

Following robjohn's technique we have

$$S=\sum_{j=0}^k\eta(k+2-j)\eta(j+2)=\sum_{j=0}^k\left(\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m^{k+2-j}}\right)\left(\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{j+2}}\right)$$

change the order of summations $$=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac{(-1)^{m+n}}{m^{k+2-j}n^{j+2}}$$

break up the middle sum $$=\sum_{m=1}^\infty\left(a_{n=m}+\sum_{n=1}^{m-1}+\sum_{n=m+1}^\infty\right)\sum_{j=0}^k\frac{(-1)^{m+n}}{m^{k+2-j}n^{j+2}}$$

pull out the terms for $$n=m$$ $$=\sum_{m=1}^\infty\sum_{j=0}^k\frac1{m^{k+4}}+\sum_{m=1}^\infty\left(\sum_{n=1}^{m-1}+\sum_{n=m+1}^\infty\right)\sum_{j=0}^k\frac{(-1)^{m+n}}{m^{k+2-j}n^{j}}$$

$$=\sum_{j=0}^k\left(\sum_{m=1}^\infty\frac1{m^{k+4}} \right)+\sum_{m=1}^\infty\left(\sum_{n=1}^{m-1}+\sum_{n=m+1}^\infty\right)\frac{(-1)^{m+n}}{m^{k+2}n^2}\left(\sum_{j=0}^k\frac{m^j}{n^j}\right)$$

$$=\sum_{j=0}^k\zeta(k+4)+\sum_{m=1}^\infty\left(\sum_{n=1}^{m-1}+\sum_{n=m+1}^\infty\right)\frac{(-1)^{m+n}}{nm^{k+2}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}$$

$$=(k+1)\zeta(k+4)+\sum_{m=1}^\infty\sum_{n=1}^{m-1}\frac{(-1)^{m+n}}{nm^{k+2}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}$$ $$+\sum_{m=1}^\infty\sum_{n=m+1}^{\infty}\frac{(-1)^{m+n}}{nm^{k+2}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}$$

By using the general change of order summation $$\sum_{m=1}^\infty \sum_{n=1}^{m-1}f(m,n)=\sum_{n=1}^\infty\sum_{m=n+1}^{\infty}f(m,n),$$

the first double sum becomes $$\sum_{m=1}^\infty\sum_{n=1}^{m-1}\frac{(-1)^{m+n}}{nm^{k+1}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}=\sum_{n=1}^\infty\sum_{m=n+1}^{\infty}\frac{(-1)^{m+n}}{nm^{k+2}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}$$

swab the variables $$n$$ and $$m$$ $$=\sum_{m=1}^\infty\sum_{n=m+1}^{\infty}\frac{(-1)^{n+m}}{mn^{k+2}(m-n)}-\frac{(-1)^{n+m}}{nm^{k+2}(m-n)}$$

$$=\sum_{m=1}^\infty\sum_{n=m+1}^{\infty}\frac{(-1)^{n+m}}{nm^{k+2}(n-m)}-\frac{(-1)^{n+m}}{mn^{k+2}(n-m)}$$

Thus, $$S=(k+1)\zeta(k+4) +2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac{(-1)^{n+m}}{nm^{k+2}(n-m)}-\frac{(-1)^{n+m}}{mn^{k+2}(n-m)}$$

reindex the inner sum $$=(k+1)\zeta(k+4) +2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{(n+m)m^{k+1}n}-\frac{(-1)^n}{m(n+m)^{k+1}n}$$ $$=(k+1)\zeta(k+4) +2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{(n+m)m^{k+2}n}-2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{m(n+m)^{k+2}n}$$

The first sum: $$S_1=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{(n+m)m^{k+2}n}=\sum_{m=1}^\infty\frac{1}{m^{k+3}}\left(\sum_{n=1}^\infty\frac{(-1)^nm}{n(n+m)}\right)$$

where $$\sum_{n=1}^\infty\frac{(-1)^nm}{n(n+m)}=\sum_{n=1}^\infty\frac{(-1)^n}{n}-\sum_{n=1}^\infty\frac{(-1)^n}{n+m}$$

$$=-\ln(2)-(-1)^n\left[\overline{H}_m-\ln(2)\right]$$

Thus, $$S_1=\sum_{m=1}^\infty\frac{1}{m^{k+3}}\left[(-1)^m \ln(2)-\ln(2)-(-1)^m\overline{H}_m\right]$$

$$=-\ln(2)\eta(k+3)-\ln(2)\zeta(k+3)-\sum_{m=1}^\infty\frac{(-1)^m\overline{H}_m}{m^{k+3}}$$

The second sum: $$S_2=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{m(n+m)^{k+2}n}=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n(n+m)}{m(n+m)^{k+3}n}$$

$$=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{m(n+m)^{k+3}}+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{(n+m)^{k+3}n}$$

swap the variables $$m$$ and $$n$$ in the first double sum and change the order of summation in the second double sum $$=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{(-1)^m+(-1)^n}{n(n+m)^{k+3}}$$

reindex the inner sum $$=\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac{(-1)^{m-n}+(-1)^n}{nm^{k+3}}$$

use $$\sum_{m=n+1}^\infty f(m)=\sum_{m=n}^\infty f(m)-f(n)$$ for the inner sum $$=\sum_{n=1}^\infty\left(\sum_{m=n}^\infty\frac{(-1)^{m-n}+(-1)^n}{nm^{k+3}}-\frac{1+(-1)^n}{n^{k+4}}\right)$$

$$=\sum_{n=1}^\infty\sum_{m=n}^\infty\frac{(-1)^{m-n}+(-1)^n}{nm^{k+3}}-\sum_{n=1}^\infty\frac{1+(-1)^n}{n^{k+4}}$$

use $$\sum_{n=1}^\infty\sum_{m=n}^\infty f(n,m)=\sum_{m=1}^\infty\sum_{n=1}^m f(n,m)$$ for the first term $$=\sum_{m=1}^\infty\sum_{n=1}^m\frac{(-1)^{m-n}+(-1)^n}{nm^{k+3}}-\zeta(k+4)+\eta(k+4)$$

$$=\sum_{m=1}^\infty\frac{1}{m^{k+3}}\left(\sum_{n=1}^m\frac{(-1)^{m-n}+(-1)^n}{n}\right)-\zeta(k+4)+\eta(k+4)$$

$$=\sum_{m=1}^\infty\frac1{m^{k+3}}\left(-(-1)^m\overline{H}_m-\overline{H}_m\right)-\zeta(k+4)+\eta(k+4)$$

$$=-\sum_{m=1}^\infty\frac{(-1)^m\overline{H}_m}{m^{k+3}}-\sum_{m=1}^\infty\frac{\overline{H}_m}{m^{k+3}}-\zeta(k+4)+\eta(k+4)$$

By combining $$S_1$$ and $$S_2$$, the term $$\sum_{m=1}^\infty\frac{(-1)^m\overline{H}_m}{m^{k+3}}$$ nicely cancels out and we have $$\sum_{j=0}^k\eta(k+2-j)\eta(j+2)$$ $$=(k+3)\zeta(k+4)-2\ln(2)[\eta(k+3)+\zeta(k+3)] -2\eta(k+4)+2\sum_{m=1}^\infty\frac{\overline{H}_m}{m^{k+3}}$$

Letting $$q=k+3$$ and reindexing $$j\mapsto j-1$$ we arrive at $$\sum_{j=1}^{q-2}\eta(q-j)\eta(j+1) =q\zeta(q+1)-2\ln(2)[\eta(q)+\zeta(q)]-2\eta(q+1)+2\sum_{m=1}^\infty\frac{\overline{H}_m}{m^q}$$

Write $$\eta(s)=(1-2^{1-s})\zeta(s)$$ we finally obtain

$$\sum_{m=1}^\infty\frac{\overline{H}_m}{m^q}=\left(1-2^{-q}-\frac{q}{2}\right)\zeta(q+1)+(2-2^{1-q})\ln(2)\zeta(q)$$ $$+\frac12\sum_{j=1}^{q-2}(1-2^{1-g+j})(1-2^{-j})\zeta(q-j)\zeta(j+1)$$

Applications

$$$$\sum_{m=1}^\infty\frac{\overline{H}_m}{m^2}=\frac32\ln(2)\zeta(2)-\frac14\zeta(3)$$$$

$$$$\sum_{m=1}^\infty\frac{\overline{H}_m}{m^3}=\frac74\ln(2)\zeta(3)-\frac5{16}\zeta(4)$$$$

$$$$\sum_{m=1}^\infty\frac{\overline{H}_m}{m^4}=\frac{15}{8}\ln(2)\zeta(4)+\frac38\zeta(2)\zeta(3)-\frac{17}{16}\zeta(5)$$$$

$$$$\sum_{m=1}^\infty\frac{\overline{H}_m}{m^5}=\frac{31}{16}\ln(2)\zeta(5)+\frac{9}{32}\zeta^2(3)-\frac{49}{64}\zeta(6)$$$$

$$$$\sum_{m=1}^\infty\frac{\overline{H}_m}{m^6}=\frac{63}{32}\ln(2)\zeta(6)+\frac{21}{32}\zeta(3)\zeta(4)+\frac{15}{32}\zeta(2)\zeta(5)-\frac{129}{64}\zeta(7)$$$$

Bonus:

By combining this generalization and $$(1)$$ from the question body and substituting the generalized euler sum we find

$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^q}=q(1-2^{-1-q})\zeta(q+1)-(2-2^{1-q})\ln(2)\zeta(q)$$ $$-\frac12\sum_{j=1}^{q-2}(2^{j+1}-1)(2^{-j}-2^{-q})\zeta(q-j)\zeta(j+1)\label{H_n/(2n+1)^q}$$

Applications

$$$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}=\frac74\zeta(3)-\frac32\ln(2)\zeta(2)$$$$

$$$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln(2)\zeta(3)$$$$

$$$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}=\frac{31}{8}\zeta(5)-\frac{15}{8}\ln(2)\zeta(4)-\frac{21}{16}\zeta(2)\zeta(3)$$$$

$$$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^5}=\frac{315}{128}\zeta(6)-\frac{31}{16}\ln(2)\zeta(5)-\frac{49}{64}\zeta^2(3)$$$$

$$$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^6}=\frac{381}{64}\zeta(7)-\frac{63}{32}\ln(2)\zeta(6)-\frac{93}{64}\zeta(2)\zeta(5)-\frac{105}{64}\zeta(3)\zeta(4)$$$$

• (+1) I was thinking of computing the alternating form of this sum. Nicely done.
– robjohn
Apr 27 '21 at 8:56
• @robjohn♦ Thank you Rob. Actually i tried your idea to prove the alternating sum and could not prove it but i found an nice way instead . I mentioned couple of your results in my book that i published few days ago ( its linked in my profile ). I hope you will like my work. Your results were really really helpful. Apr 27 '21 at 15:40

The value of the series may be extracted from Theorem $$1$$ of the preprint A simple strategy of calculating two alternating harmonic series generalizations. More precisely, we have

Let $$m\ge2$$ be a positive integer. The following equalities hold: $$\begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x \end{equation*}$$ $$\begin{equation*} \small =\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr), \end{equation*}$$ where $$H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$$ represents the $$n$$th generalized harmonic number of order $$m$$ and $$\zeta$$ denotes the Riemann zeta function.

One may also check https://math.stackexchange.com/q/3236584.

A note: the series mentioned by omegadot (from a paper by R. Sitaramachandrarao, "A formula of S. Ramanujan," Journal of Number Theory, 25, 1-19 (1987). See Theorem 3.5 on page 9) seems to be strongly related to the series above if we look at their integral representations, which will be mentioned in the next version of the paper.