Two challenging sums $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}$ 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".
 A: Both series are calculated by simple real techniques in the book, (Almost) Impossible Integrals, Sums, and Series,
$$a) \ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^3}=\frac{5}{8}\zeta(2)\zeta(3)-\frac{11}{32}\zeta(5);$$
$$b) \ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^3}$$
$$=\frac{2}{15}\log^5(2)-\frac{11}{8}\zeta(2)\zeta(3)-\frac{19}{32}\zeta(5)+\frac{7}{4}\log^2(2)\zeta(3)-\frac{2}{3}\log^3(2)\zeta(2)$$
$$+4\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)+4\operatorname{Li}_5\left(\frac{1}{2}\right).$$
A: 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

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

A: 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.
