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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 $a>4$.

Thanks in advance.

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 $\pi^a$.

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\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 $S_2$.

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  • $\begingroup$ @ 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. $\endgroup$ – Dr. Wolfgang Hintze Aug 2 '19 at 15:47
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    $\begingroup$ @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. $\endgroup$ – Ali Shadhar Aug 2 '19 at 21:11
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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 $a-k<a$ 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 $<a-k$ 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 $n=-m$.

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  • $\begingroup$ (+1) amazing proof. can you give details about how you got these coefficients ? $\endgroup$ – Ali Shadhar Aug 23 at 22:59

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