Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$

Question

where $H_n$ is the harmonic number and can be defined as:

$H_n=1+\frac12+\frac13+...+\frac1n$

$H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$

these two sums are already solved by Cornel using summation manipulation and can be also found in his newly released book "(almost) impossible integrals, sums and series ".

I was able to evaluate them using integration and some harmonic identities.

\begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}&=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac23\ln^32\zeta(2)+\frac74\ln^22\zeta(3)\\&\quad-\frac{15}{16}\zeta(2)\zeta(3)-\frac{23}8\zeta(5)+\frac2{15}\ln^52 \end{align} \begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}&=-6\operatorname{Li}_5\left(\frac12\right)-6\ln2\operatorname{Li}_4\left(\frac12\right)+\ln^32\zeta(2)-\frac{21}{8}\ln^22\zeta(3)\\&\quad+\frac{27}{16}\zeta(2)\zeta(3)+\frac94\zeta(5)-\frac15\ln^52 \end{align}

The point of posting these two sums is to use them as a reference in our solutions if needed.

Answer 1

To calculate these two sums, we are going to establish two relations and solve them by elimination.

To establish the first relation, we use $\displaystyle I=\int_0^1\frac{\ln^4(1+x)+6\ln^2(1-x)\ln^2(1+x)}x\ dx=\frac{21}4\zeta(5)\tag{1}$

which was proved by Khalef Ruhemi (unfortunately he is not an MSE user).

The proof as follows: using the algebraic identity $b^4+6a^2b^2=\frac12(a-b)^4+\frac12(a+b)^4-a^4$

with $a=\ln(1-x)$ and $b=\ln(1+x)$, divide both sides by $x$ then integrate, we get

$$\eqalign{ I&=\frac12\underbrace{\int_0^1\frac1x{\ln^4\left(1-x\over1+x\right)}\ dx}_{{1-x\over1+x}=y}+\underbrace{\frac12\int_0^1\frac{\ln^4(1-x^2)}x\ dx}_{x^2=y}-\int_0^1\frac{\ln^4(1-x)}x\ dx\\ &=\int_0^1{\ln^4x\over1-x^2}+\frac14\int_0^1\frac{\ln^4(1-x)}x\ dx-\int_0^1\frac{\ln^4(1-x)}x\ dx\\ &=\frac12\int_0^1{\ln^4x\over1-x}+\frac12\int_0^1{\ln^4x\over1+x}-\frac34\underbrace{\int_0^1\frac{\ln^4(1-x)}x\ dx}_{1-x=y}\\ &=\frac12\int_0^1{\ln^4x\over1+x}\ dx+\frac14\int_0^1{\ln^4x\over1-x}\ dx=\frac12\left(\frac{45}2\zeta(5)\right)+\frac14(24\zeta(5))=\frac{21}4\zeta(5) }$$

---

On the other hand, $\quad\displaystyle I=\underbrace{\int_0^1\frac{\ln^4(1+x)}x\ dx}_{I_1}+6\int_0^1\frac{\ln^2(1-x)\ln^2(1+x)}x\ dx$

Using $\ln^2(1+x)=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}n-\frac1{n^2}\right)x^n\ $ for the second integral, we get

$$\eqalign{ I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}n-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^2(1-x)\ dx\\ &=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}n-\frac1{n^2}\right)\left(H_n^2+H_n^{(2)}\over n\right)\\ &=I_1+12\sum_{n=1}^\infty(-1)^n\left(H_n^3+H_nH_n^{(2)}\over n^2\right)-12\sum_{n=1}^\infty(-1)^n\left(H_n^2+H_n^{(2)}\over n^3\right)\tag{2} }$$ From $(1)$ and $(2)$, we get

$$\boxed{\small{R_1=\sum_{n=1}^\infty{(-1)^nH_n^3\over n^2}+\sum_{n=1}^\infty{(-1)^nH_nH_n^{(2)}\over n^2}=\frac7{16}\zeta(5)+\sum_{n=1}^\infty{(-1)^nH_n^2\over n^3}+\sum_{n=1}^\infty{(-1)^nH_n^{(2)}\over n^3}-\frac1{12}I_1}}$$

and the first relation is established.

---

To get the second relation, we need to use the Stirling number formula (check here) $${\ln^k(1-x)\over k!}=\sum_{n=k}^\infty(-1)^k{n\brack k}{x^n\over n!}$$ letting $k=4$ and using $\displaystyle{n\brack4}=\frac1{3!}(n-1)!\left[\left(H_{n-1}\right)^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right],$ we get $$\frac14\ln^4(1-x)=\sum_{n=1}^\infty{x^{n+1}\over n+1}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

differentiate both sides with respect to $x$, we get $$-{\ln^3(1-x)\over1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Now replace $x$ with $-x$ then multiply both sides by $\ln x\over x$ and integrate, we get $$-\sum_{n=1}^\infty(-1)^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1x^{n-1}\ln x\ dx=\int_0^1{\ln^3(1+x)\ln x\over x(1+x)}\ dx$$ $$\eqalign{ \sum_{n=1}^\infty {(-1)^n\over n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)&=\int_0^1\frac{\ln^3(1+x)\ln x}x\ dx-\underbrace{\int_0^1{\ln^3(1+x)\ln x\over1+x}\ dx}_{IBP}\\ &=\int_0^1\frac{\ln^3(1+x)\ln x}x\ dx+\frac14I_1 }$$ Rearranging the terms, we get $$\boxed{R_2=\sum_{n=1}^\infty{(-1)^nH_n^3\over n^2}-3\sum_{n=1}^\infty{(-1)^nH_nH_n^{(2)}\over n^2}=\int_0^1\frac{\ln^3(1+x)\ln x}x-2\sum_{n=1}^\infty{(-1)^nH_n^{(3)}\over n^2}+\frac14I_1}$$ and the second relation is established.

---

Now we are ready to calculate the first sum. $$\eqalign{ \sum_{n=1}^\infty{(-1)^nH_n^3\over n^2}&=\frac{3R_1+R_2}4\\ &=\frac34\sum_{n=1}^\infty{(-1)^nH_n^2\over n^3}+\frac34\sum_{n=1}^\infty{(-1)^nH_n^{(2)}\over n^3}-\frac12\sum_{n=1}^\infty{(-1)^nH_n^{(3)}\over n^2}\\ &\quad+\frac14\int_0^1\frac{\ln x\ln^3(1+x)}x\ dx+{21\over64}\zeta(5) }$$ the closed form of the first and second sum can be found here and the closed form of the third sum is evaluated here. as for the integral, I evaluated it here. by combining these results, we get our closed form.

and the second sum. $$\sum_{n=1}^\infty{(-1)^nH_nH_n^{(2)}\over n^2}=\frac{R_1-R_2}4$$ $$\small{=\frac14\sum_{n=1}^\infty{(-1)^nH_n^2\over n^3}+\frac14\sum_{n=1}^\infty{(-1)^nH_n^{(2)}\over n^3}+\frac12\sum_{n=1}^\infty{(-1)^nH_n^{(3)}\over n^2}-\frac14\int_0^1\frac{\ln x\ln^3(1+x)}x\ dx-\frac1{12}I_1+\frac7{64}\zeta(5)}$$ let's calculate $I_1$; by setting $\frac1{1+x}=y$, we get $$\eqalign{ I_1&=\int_0^1\frac{\ln^4(1+x)}x=\int_{1/2}^1\frac{\ln^4x}x\ dx+\int_{1/2}^1{\ln^4x\over1-x}\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\left({24\over n^5}-{24\over n^52^n}-{24\ln2\over n^42^n}-{12\ln^22\over n^32^n}-{4\ln^32\over n^22^n}-{\ln^42\over n2^n}\right)\\ &=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right) }$$ by combining the result of $I_1$ with the results we used in our first sum, we get the closed form of the second sum.

---

UPDATE:

The identity used above:

$$-{\ln^3(1-x)\over1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

can also be proved this way.

Answer 2

Different approach

First let define

$$M=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$$

$$N=\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$$

$$P=\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}$$

I am going to establish two relations and solve them as a system of equations.

---

First relation:

From here we have

$$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$

Multiply both sides by $\frac{(-1)^n}{n}$ then $\sum_{n=1}^\infty$ we have

$$M+3N+2P=\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{(-x)^{n-1}}{n}dx=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx\tag1$$

---

Second relation:

From here we have

$$\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)x^n=-\frac{\ln^3(1-x)}{1-x}$$

Replace $x$ with $-x$ and multiply both sides by $-\frac{\ln x}{x}$ then $\int_0^1$ and use $-\int_0^1 x^{n-1}\ln xdx=\frac{1}{n^2}$ we get

$$M-3N+2P=\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}dx\tag2$$

---

So

$$M=\frac{(1)+(2)}{2}=\frac12\color{blue}{\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx}+\frac12\color{red}{\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}dx}-2P$$

the first integral is nicely calculated by @Song here

$$\color{blue}{\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx}=6\operatorname{Li}_5\left(\frac12\right)+6\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{16}\zeta(5)-\frac{21}{8}\zeta(2)\zeta(3)\quad+\frac{21}8\ln^22\zeta(3)-\ln^32\zeta(2)+\frac15\ln^52$$

For the second integral,

$$\color{red}{\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}dx}=\int_0^1\frac{\ln^3(1+x)\ln x}{x}dx-\underbrace{\int_0^1\frac{\ln^3(1+x)\ln x}{1+x}dx}_{IBP}$$

$$=\int_0^1\frac{\ln^3(1+x)\ln x}{x}dx+\frac14\int_0^1\frac{\ln^4(1+x)}{x}dx$$

where the first part is calculated here

$$\int_0^1\frac{\ln^3(1+x)\ln x}{x}dx=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{99}{16}\zeta(5)+3\zeta(2)\zeta(3)\\-\frac{21}4\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52$$

and the second part can be calculated through the generalization

$$\int_0^1\frac{\ln^n(1+x)}{x}dx=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$

which gives

$$\small{\int_0^1\frac{\ln^4(1+x)}{x}dx=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)}$$

combine the two parts we have

$$\color{red}{\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}dx}=-18\operatorname{Li}_5\left(\frac12\right)-18\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{195}{16}\zeta(5)+3\zeta(2)\zeta(3)\\-\frac{63}8\ln^22\zeta(3)+3\ln^32\zeta(2)-\frac35\ln^52$$

For the sum $P$, its already calculated here

$$P=\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$

Combine the results of the blue integral, red integral and $P$ we get

$$M=-6\operatorname{Li}_5\left(\frac12\right)-6\ln2\operatorname{Li}_4\left(\frac12\right)+\ln^32\zeta(2)-\frac{21}{8}\ln^22\zeta(3)+\frac{27}{16}\zeta(2)\zeta(3)+\frac94\zeta(5)-\frac15\ln^52$$

---

And

$$N=\frac{(1)-(2)}{6}=\frac16\color{blue}{\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx}-\frac16\color{red}{\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}dx}$$

$$=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac23\ln^32\zeta(2)+\frac74\ln^22\zeta(3)-\frac{15}{16}\zeta(2)\zeta(3)-\frac{23}8\zeta(5)+\frac2{15}\ln^52$$