# Compute $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}$ and $\sum_{n=1}^\infty\frac{H_n^2}{n^7}$

How to prove that

$$S_1=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}=7\zeta(2)\zeta(7)+2\zeta(3)\zeta(6)+4\zeta(4)\zeta(5)-\frac{35}{2}\zeta(9)\ ?$$ $$S_2=\sum_{n=1}^\infty\frac{H_n^2}{n^7}=-\zeta(2)\zeta(7)-\frac72\zeta(3)\zeta(6)+\frac13\zeta^3(3)-\frac{5}{2}\zeta(4)\zeta(5)+\frac{55}{6}\zeta(9)\ ?$$ where $$H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$$ is the $$n$$th generalized harmonic number of order $$p$$.

I came across these two sums while working on an tough one of wight 9 and I managed to find these two results but I don't think my solution is a good approach as it's pretty long and involves tedious calculations, so I am seeking different methods if possible. I am much into new ideas. All approaches are appreciated though.

By the way, do we have a generalization for $$\displaystyle\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^a}$$, for odd $$a$$? Note that there is no closed form for even $$a>4$$.

Note: You can find these two results on Wolfram Alpha here and here respectively but I modified it a little bit as I like it expressed in terms of $$\zeta(a)$$ instead of $$\pi^a$$.

\begin{align} S_1&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}=\sum_{n=1}^\infty\frac1{n^7}\left(\zeta(2)-\sum_{k=1}^\infty\frac1{(n+k)^2}\right)\\ &=\zeta(2)\zeta(7)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac{1}{n^7(n+k)^2}\right)\\ &\small{=\zeta(2)\zeta(7)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac{7}{k^8}\left(\frac1n-\frac1{n+k}\right)-\frac{6}{k^7n^2}-\frac{1}{k^7(n+k)^2}+\frac{5}{k^6n^3}-\frac{4}{k^5n^4}+\frac{3}{k^4n^5}-\frac{2}{k^3n^6}+\frac{1}{k^2n^7}\right)}\\ &\small{=\zeta(2)\zeta(7)-\sum_{k=1}^\infty\left(\frac{7H_k}{k^8}-\frac{6\zeta(2)}{k^7}-\frac1{k^7}\left(\zeta(2)-H_k^{(2)}\right)+\frac{5\zeta(3)}{k^6}-\frac{4\zeta(4)}{k^5}+\frac{3\zeta(5)}{k^4}-\frac{2\zeta(6)}{k^3}+\frac{\zeta(7)}{k^2}\right)}\\ &=\zeta(2)\zeta(7)-7\sum_{k=1}^\infty\frac{H_k}{k^8}+6\zeta(2)\zeta(7)-S_1-3\zeta(3)\zeta(6)+\zeta(4)\zeta(5)\\ 2S_1&=7\zeta(2)\zeta(7)-3\zeta(3)\zeta(6)+\zeta(4)\zeta(5)-7\sum_{k=1}^\infty\frac{H_k}{k^8} \end{align}

and by substituting $$\displaystyle\sum_{k=1}^\infty\frac{H_k}{k^8}=5\zeta(9)-\zeta(2)\zeta(7)-\zeta(3)\zeta(6)-\zeta(4)\zeta(5)$$, we get the closed form of $$S_1$$.

From here, we have

$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$

Divide both sides by $$n^6$$ then sum both sides from $$n=1$$ to $$\infty$$ to get

\begin{align} S_2+S_1&=\sum_{n=1}^\infty\frac{H_n^2}{n^7}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{x^n}{n^6}\ dx\\ &=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_6(x)}{x}\ dx,\quad \left\{\color{red}{\text{use}\ \ln^2(1-x)=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)x^n}\right\}\\ &=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1} \operatorname{Li}_6(x)\ dx,\quad \left\{\color{red}{\text{apply integration by parts}}\right\}\\ &=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)\left(\frac{\zeta(6)}{n}-\frac{\zeta(5)}{n^2}+\frac{\zeta(4)}{n^3}-\frac{\zeta(3)}{n^4}+\frac{\zeta(2)}{n^5}-\frac{H_n}{n^6}\right)\\ 3S_2+S_1&=2\sum_{n=1}^\infty\frac{H_n}{n^8}+2\zeta(6)\sum_{n=1}^\infty\frac{H_n}{n^2}-2\zeta(5)\sum_{n=1}^\infty\frac{H_n}{n^3}+2\zeta(4)\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &\quad-2\zeta(3)\sum_{n=1}^\infty\frac{H_n}{n^5}+2\zeta(2)\sum_{n=1}^\infty\frac{H_n}{n^6}-2\zeta(2)\zeta(7)\tag{1} \end{align}

From Euler's identity, we can obtain the following results:

$$\sum_{n=1}^\infty\frac{H_n}{n^2}=2\zeta(3)$$ $$\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$$ $$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$ $$\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$ $$\sum_{n=1}^\infty\frac{H_n}{n^6}=4\zeta(7)-\zeta(2)\zeta(5)-\zeta(3)\zeta(4)$$ $$\sum_{k=1}^\infty\frac{H_k}{k^8}=5\zeta(9)-\zeta(2)\zeta(7)-\zeta(3)\zeta(6)-\zeta(4)\zeta(5)$$

By plugging these results along with that of $$S_1$$ in $$(1)$$, we get the closed form of $$S_2$$.

• @ Ali Shather Considering the impressive amount of similar problems/answers you have posted I would find it helpful if you would embed your abundant results into the literature on the subject. Especially point out what is already known and what is really new. Aug 2, 2019 at 15:47
• @Dr.WolfgangHintze thank you I think most results I posted are already known in the literature but I am just presenting new solutions and seeking different ideas. Still, even if I get a new result, its not easy to determine if its known or not. I will post a nice problem related to your suggestion in the comment. Aug 2, 2019 at 21:11

Your method can be carried out in full generality as long as the power $$a$$ is odd. Starting with $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^a} = \zeta(2)\zeta(a)-\sum_{n,m=1}^\infty \frac{1}{n^a(n+m)^2} \, ,$$ we can generally decompose the fraction in the second term into partial fractions $$\frac{1}{n^a(n+m)^2}=\sum_{k=1}^a \frac{A_k}{n^k}+\frac{B_1}{n+m} + \frac{B_2}{(n+m)^2} \, . \tag{1}$$ By the method of residues these coefficients are given by $$A_k=\frac{(-1)^{a-k}(a-k+1)}{m^{a-k+2}} \\ B_1=\frac{a\,(-1)^a}{m^{a+1}} \\ B_2=\frac{(-1)^a}{m^a}$$ and hence $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^a} = \zeta(2)\zeta(a)-\sum_{n,m=1}^\infty \frac{a \, (-1)^{a-1}}{m^{a+1}} \left(\frac{1}{n} - \frac{1}{m+n} \right) \\ -\sum_{n,m=1}^\infty \sum_{k=2}^a \frac{(-1)^{a-k}(a-k+1)}{n^k m^{a-k+2}} - \sum_{n,m=1}^\infty \frac{(-1)^a}{m^a (m+n)^2} \\ = \zeta(2)\zeta(a) + a \, (-1)^a \sum_{m=1}^\infty \frac{H_m}{m^{a+1}} \\ - \sum_{k=2}^a (-1)^{a-k} (a-k+1) \zeta(k)\zeta(a-k+2) - (-1)^a \left(\zeta(2)\zeta(a) - \sum_{m=1}^\infty \frac{H_m^{(2)}}{m^a} \right) \, .$$ For even $$a$$ you will get an identity similar to that of Euler. For odd $$a$$ you can solve for the LHS $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^a} = \zeta(2)\zeta(a) - \frac{a}{2} \sum_{m=1}^\infty \frac{H_m}{m^{a+1}} + \frac{1}{2} \sum_{k=2}^a (-1)^{k} (a-k+1) \zeta(k)\zeta(a-k+2) \, .$$

You can simplify it further by substituting Euler's identity for the middle term.

The coefficients in (1) are obtained as follows. For $$A_k$$ multiply (1) by $$n^a$$ and derive both sides with respect to $$n$$ exactly $$a-k$$ times. Finally set $$n=0$$. This way the LHS will yield $$\frac{(-1)^{a-k} \, (a-k+1)!}{m^{a-k+2}}$$. The $$B$$-terms on the RHS will vanish, because $$n^a$$ is derived at most $$a-k times, which leaves at least one power in $$n$$ which is set to zero. For the $$A$$-terms the monomials $$n^{a-k}$$ are successively derived and those with power $$ vanish immediately, while those with power $$>a-k$$ will vanish upon setting $$n=0$$. The term with power $$a-k$$ gives $$(a-k)! A_k$$.

The $$B$$ coefficients are obtained in the same way by multiplying (1) with $$(n+m)^2$$, deriving zero or one times and setting $$n=-m$$.

• (+1) amazing proof. can you give details about how you got these coefficients ? Aug 23, 2020 at 22:59