# Prove $\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)$

How to prove this advanced harmonic series of weight 6?

$$S=\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)$$ where $$H_k^{(p)}=1+\frac1{2^p}+\cdots+\frac1{k^p}$$ is the $$k$$th generalized harmonic number of order $$p$$.

This result can be found in the book Almost impossible integrals, sums and series page $$414-419$$ using pure series manipulations but can be done in different ways?

I will post my approach soon.

Using the fact that $$\sum_{n=1}^\infty H_n^{(3)}x^n=\frac{\operatorname{Li}_3(x)}{1-x}$$ Divide both sides by $$x$$ then integrate, we get

\begin{align} \sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}&=\int \frac{\operatorname{Li}_3(x)}{x(1-x)}\ dx=\int \frac{\operatorname{Li}_3(x)}{x}\ dx+\underbrace{\int \frac{\operatorname{Li}_3(x)}{1-x}\ dx}_{IBP}\\ &=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)+\int\frac{\ln(1-x)\operatorname{Li}_2(x)}{x}\ dx\\ &=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)+c \end{align} Set $$x=0$$, we get $$c=0$$.

Therefore

$$\sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)\tag{1}$$

Now multiply both sides of $$(1)$$ by $$-\frac{\ln(1-x)}{x}$$ then integrate, we get \begin{align} S&=\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}\int_0^1 -x^{n-1}\ln(1-x)\ dx=\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}\left(\frac{H_n}{n}\right)=\sum_{n=1}^\infty \frac{H_n^{(3)}H_n}{n^2}\\ &=-\int_0^1\frac{\ln(1-x)\operatorname{Li}_4(x)}{x}\ dx+\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_3(x)}{x}\ dx+\frac12\int_0^1\frac{\ln(1-x)\operatorname{Li}^2_2(x)}{x}\ dx\\ &=-\sum_{n=1}^\infty\frac1{n^4}\int_0^1 x^{n-1}\ln(1-x)\ dx+\sum_{n=1}^\infty\frac1{n^3}\int_0^1 x^{n-1}\ln^2(1-x)\ dx-\frac16\operatorname{Li}^3_2(1)\\ &=-\sum_{n=1}^\infty\frac1{n^4}\left(-\frac{H_n}{n}\right)+\sum_{n=1}^\infty\frac1{n^3}\left(\frac{H^2_n}{n}+\frac{H_n^{(2)}}{n}\right)-\frac16\zeta^3(2)\\ &=\sum_{n=1}^\infty \frac{H_n}{n^5}+\sum_{n=1}^\infty \frac{H^2_n}{n^4}+\sum_{n=1}^\infty \frac{H^{(2)}_n}{n^4}-\frac{35}{48}\zeta(6) \end{align} By substituting: $$\sum_{n=1}^\infty \frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$

$$\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$

$$\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\zeta^2(3)-\frac13\zeta(6)$$

we get the closed form $$\boxed{S=\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)}$$

Note: The first sum can be found using Euler identity, the second sum is evaluated here and the third sum is here.

Here is my slight variation on how I go about things. As you will see, the ideas used and results found are almost the same as yours.

I will make use of result (1) you kindly provide in your answer, namely

$$\sum_{n = 1}^\infty \frac{H^{(3)}_n x^n}{n} = \operatorname{Li}_4 (x) - \ln (1 - x) \operatorname{Li}_3 (x) - \frac{1}{2} \operatorname{Li}^2_2 (x)\tag1$$

Since $$\int_0^1 x^{n - 1} \ln (1 - x) \, dx = -\frac{H_n}{n},$$ one can express the sum as $$\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^2} = \sum_{n = 1}^\infty \frac{H^{(3)}_n}{n} \cdot \frac{H_n}{n} = -\int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{H^{(3)}_n x^n}{n} \, dx\tag2$$ Substituting (1) into (2) leads to \begin{align} \sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^2} &= - \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_4 (x)}{x} \, dx + \int_0^1 \frac{\ln^2 (1 - x) \operatorname{Li}_3 (x)}{x} \, dx\\ & \qquad + \frac{1}{2} \int_0^1 \frac{\ln (1 - x) \operatorname{Li}^2_2 (x)}{x} \, dx\\ &= -I_1 + I_2 + \frac{1}{2} I_3. \end{align}

The first integral $$I_1$$

\begin{align} I_1 &= \underbrace{\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_4 (x)}{x} \, dx}_{IBP \,\, 3 \,\, \text{times}}\\ &= -\sum_{n = 1}^\infty \left [\frac{\zeta (4)}{n^2} - \frac{\zeta (3)}{n^3} + \frac{\zeta (2)}{n^4} + \frac{1}{n^4} \int_0^1 x^{n - 1} \ln (1 - x) \, dx \right ]\\ &= -\zeta (4) \sum_{n = 1}^\infty \frac{1}{n^2} + \zeta (3) \sum_{n = 1}^\infty \frac{1}{n^3} - \zeta (2) \sum_{n = 1}^\infty \frac{1}{n^4} + \sum_{n = 1}^\infty \frac{H_n}{n^5}\\ &= -\zeta (4) \zeta (2) + \zeta^2 (3) - \zeta (2) \zeta (4) + \frac{7}{4} \zeta (6) - \frac{1}{2} \zeta^2 (3)\\ &= \frac{1}{2} \zeta^2 (3) - \frac{7}{4} \zeta (6), \end{align} where the results $$\sum_{n = 1}^\infty \frac{H_n}{n^5} = \frac{7}{4} \zeta (6) - \frac{1}{2} \zeta^2 (3) \quad \text{and} \quad \zeta (2) \zeta (4) = \frac{7}{4} \zeta (6),$$ were used.

The second integral $$I_2$$

In this question here I showed that $$I_2 = 2 \zeta (3) \sum_{n = 1}^\infty \frac{H_n}{n^2} - 2 \zeta^2 (3) - 2 \zeta (2) \sum_{n = 1}^\infty \frac{H_n}{n^3} + 2 \zeta (2) \zeta (4) + 2 \sum_{n = 1}^\infty \frac{H^2_n}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^5}.$$ All four of the sums appearing in the above expression for $$I_2$$ are known. The first, second, and fourth sums are standard Euler sums while a proof of the third sum can be found here. The results are: \begin{align} \sum_{n = 1}^\infty \frac{H_n}{n^2} &= 2 \zeta (2)\\ \sum_{n = 1}^\infty \frac{H_n}{n^3} &= \frac{5}{4} \zeta (4)\\ \sum_{n = 1}^\infty \frac{H_n}{n^5} &= -\frac{1}{2} \zeta^2 (3) + \frac{7}{4} \zeta (6)\\ \sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \frac{97}{24} \zeta (6) - 2 \zeta^2 (3), \end{align} Thus $$I_2 = -\zeta^2 (3) + \frac{89}{24} \zeta (6).$$

The third integral $$I_3$$

\begin{align} I_3 &= \underbrace{\int_0^1 \frac{\ln (1 - x) \operatorname{Li}^2_2 (x)}{x} \, dx}_{IBP}\\ &= -\operatorname{Li}^3_2(1) - 2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}^2_2 (x)}{x} \, dx\\ \Rightarrow I_3 &= -\frac{1}{3} \zeta^3 (2) = -\frac{35}{24} \zeta (6). \end{align}

The main sum

Combining the results found for the above three integrals, for the sum one has $$\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^2} = \frac{227}{48} \zeta (6) - \frac{3}{2} \zeta^2 (3),$$ as required.

• Nice. Regarding $I_3$ , the derivative of $Li_2(x)$ is already there so no need for IBP. – Ali Shadhar Jul 20 '19 at 3:39