3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.