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

Question

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

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

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

again, my goal of posting these two challenging sums is to use them as a reference.

I will provide my solutions soon.

I would like to mention that these two sums can also be found in Cornel's book " almost impossible integrals, sums, and series".

Answer 1

Both series are calculated by simple real techniques in the book, (Almost) Impossible Integrals, Sums, and Series,

$$a) \ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^3}=\frac{5}{8}\zeta(2)\zeta(3)-\frac{11}{32}\zeta(5);$$

$$b) \ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^3}$$ $$=\frac{2}{15}\log^5(2)-\frac{11}{8}\zeta(2)\zeta(3)-\frac{19}{32}\zeta(5)+\frac{7}{4}\log^2(2)\zeta(3)-\frac{2}{3}\log^3(2)\zeta(2)$$ $$+4\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)+4\operatorname{Li}_5\left(\frac{1}{2}\right).$$

Answer 2

Using the fact that $\displaystyle \sum_{n=1}^\infty x^nH_n^{(2)}=\frac{\operatorname{Li}_2(x)}{1-x}$

Replace $x$ with $-x$ then multiply both sides by $\ln^2x$ and integrate, we get \begin{align} S&=\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1x^{n}\ln^2x\ dx=2\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{(n+1)^3}=\underbrace{\int_0^1\frac{\ln^2x\operatorname{Li}_2(-x)}{1+x}\ dx}_{IBP}\\ &=\int_0^1\frac{\ln^2x \ln^2(1+x)}{x}\ dx-2\int_0^1\frac{\ln x\ln(1+x)\operatorname{Li}_2(-x)}{x}\ dx\\ &=I_1-2I_2 \end{align} Lets evaluate the first integral and using $\quad \ln^2(1+x)=2\sum_{n=1}^\infty (-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n,\quad $ we get \begin{align} I_1&=2\sum_{n=1}^\infty (-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln^2x\ dx\\ &=2\sum_{n=1}^\infty (-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(\frac{2}{n^3}\right)\\ &=4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-4\sum_{n=1}^\infty\frac{(-1)^n}{n^5}\\ &=4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{15}{4}\zeta(5) \end{align} to evaluate the second integral, apply IBP , we get \begin{align} I_2&=\left.-\frac12\operatorname{Li}_2^2(-x)\ln x\right|_0^1+\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ &=\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ \end{align} I proved here $\quad \displaystyle \int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx=\frac58\zeta(2\zeta(3)+\frac78\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$

Collecting these two integrals and using $\quad \displaystyle \sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3),\quad$ we get $$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{(n+1)^3}=\frac9{16}\zeta(5)+\frac18\zeta(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$ but $$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{(n+1)^3}=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{n^3}-\frac{15}{16}\zeta(5)$$ Thus $$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{n^3}=\frac32\zeta(5)+\frac18\zeta(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$ Plugging $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$ gives

$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{n^3}=\frac58\zeta(2)\zeta(3)-\frac{11}{32}\zeta(5)$$

Answer 3

Lets calculate the second sum and using the identity $\quad \displaystyle \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$

Replace $x$ with $-x$, then multiply both sides by $\ln^2x$ and integrate, we get \begin{align} I&=\int_0^1\frac{\ln^2x\ln^2(1+x)}{1+x}\ dx=\sum_{n=1}^\infty (-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^n\ln^2x\ dx\\ &=2\sum_{n=1}^\infty (-1)^n\frac{H_n^2-H_n^{(2)}}{(n+1)^3}=-2\sum_{n=1}^\infty (-1)^n\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n^3}\\ &=-2\sum_{n=1}^\infty (-1)^n\left(\frac{H_n^2}{n^3}-\frac{H_n^{(2)}}{n^3}-2\frac{H_n}{n^4}+\frac{2}{n^5}\right)\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^2}{n^3}-2\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{n^3}+4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{15}4\zeta(5) \end{align} we have already proved $$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{n^3}=\frac32\zeta(5)+\frac18\zeta(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$

Thus $$I=\frac34\zeta(5)-\frac14\zeta(2)\zeta(3)+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+2\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^2}{n^3}\tag{1}$$

applying IBP for the integral, we get $\quad \displaystyle I=-\frac23\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx$

I managed here to prove \begin{align} \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)\\ &\quad-\frac{21}4\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52 \end{align} giving us $$I=8\operatorname{Li}_5\left(\frac12\right)+8\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{33}{8}\zeta(5)-2\zeta(2)\zeta(3)+\frac72\ln^22\zeta(3)-\frac43\ln^32\zeta(2)+\frac4{15}\ln^52$$ Plugging the value of $I$ in $(1)$ along with the value of $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$, we get

\begin{align} \sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^2}{n^3}&=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{19}{32}\zeta(5)-\frac{11}8\zeta(2)\zeta(3)\\ &\quad+\frac74\ln^22\zeta(3)-\frac23\ln^32\zeta(2)+\frac2{15}\ln^52 \end{align}

Answer 4

A much easier approach:

By Cauchy product we have

$$-\ln(1-x)\operatorname{Li}_2(x)=\sum_{n=1}^\infty\left(\frac{2H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{3}{n^3}\right)x^n$$

replace $x$ with $-x$ then multiply both sides by $-\frac{\ln x}{x}$ and integrate between $0$ and $1$ plus use the fact that $\int_0^1-x^{n-1}\ln x\ dx=\frac1{n^2}$ we get

$$2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-3\operatorname{Li}_5(-1)=\int_0^1\frac{\ln(1+x)\operatorname{Li}_2(-x)\ln x}{x}dx$$

$$\overset{IBP}{=}\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}dx=\frac{5}{16}\zeta(2)\zeta(3)+\frac{7}{16}\sum_{n=1}^\infty\frac{H_n}{n^4}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$

where the last result follows from this solution, check Eq$(3)$.

rearrange to get

$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}=\frac{5}{16}\zeta(2)\zeta(3)-\frac{45}{16}\zeta(5)+\frac{7}{16}\sum_{n=1}^\infty\frac{H_n}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$

substitute $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$ and $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$, we get

$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n^{(2)}}{n^3}=\frac{11}{32}\zeta(5)-\frac58\zeta(2)\zeta(3)$$

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Bonus:

Again, by Cauchy product we have

$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)x^n$$

set $x=-1$ and substitute the result of $\sum_{n=1}^\infty\frac{(-1)^{n}H_n^{(2)}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^4}$ we get

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

Or it can be found here.