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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".

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    $\begingroup$ (+1) for the question. Love to see more solutions. $\endgroup$ – user97357329 Jun 5 at 13:13
  • $\begingroup$ @user97357329 I will try to post today. $\endgroup$ – Ali Shather Jun 5 at 18:25
  • $\begingroup$ Possible repetitions math.stackexchange.com/questions/2169507/… $\endgroup$ – Dr. Wolfgang Hintze Jun 6 at 6:40
  • $\begingroup$ @Dr. Wolfgang Hintze nice link but I dont think it helps us here to solve our two sums. if you think there is a helpful formula in the link you provided, would you spot it for us? $\endgroup$ – Ali Shather Jun 6 at 11:05
  • $\begingroup$ @ Ali Shather It is not only nice but - for the linear case - very comprehensive, and it would have been nice and good practice if you would have quoted previous work of Prztemo and point out to the reader what is really new. Also you might wish to discover your linear formula by yourself in the solution math.stackexchange.com/a/2264045/198592 $\endgroup$ – Dr. Wolfgang Hintze Jun 6 at 12:19
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Both series are calculated by simple real techniques in the book, (Almost) Impossible Integrals, Sums, and Series,

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

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

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  • $\begingroup$ Yes I forgot to mention that. And you can find tougher ones in that book. Anyway i solved these two sums and more long time ago and in a different approach. $\endgroup$ – Ali Shather Jun 5 at 17:48
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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 the well known result $\quad \displaystyle \sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$

finally we get

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

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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} which follows that $$I=8\operatorname{Li}_5\left(\frac12\right)+8\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{33}{8}\zeta(5)-2\zeta(2)\zeta(3)+\frac72\ln^22\zeta(3)-\frac43\ln^32\zeta(2)+\frac4{15}\ln^52$$ 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}

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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

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


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.

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