Problem: Closed form of $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2}$
We have seen how to deal with the Au-Yeung Series in three different ways in Closed form of Enrico Au-Yeung Series. However alternating versions (A) of Euler Sums often turn out to be trickier to deal with than the (H) counterpart.
Solution: We begin with the key identity:
$$\begin{align}\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}\frac{H_{k+1}^{(3)}}{k+1} &= \frac{1}{n+1}\sum_{k=0}^{n} (-1)^k \binom{n+1}{k+1}\sum\limits_{j=0}^{k}\frac{1}{(j+1)^3}\tag{1}\\ &= \frac{1}{n+1}\sum_{j=0}^{n}\frac{1}{(j+1)^3} \sum\limits_{k=j}^{n}(-1)^k \binom{n+1}{k+1}\tag{2}\\ &= \frac{1}{n+1}\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\frac{1}{(j+1)^3}\tag{3} \\ &= \frac{1}{(n+1)^2}\sum_{j=0}^{n}(-1)^{j} \binom{n+1}{j+1}\frac{1}{(j+1)^2} \tag{4}\\ &= \frac{1}{(n+1)^2}\sum_{j=0}^{n}(-1)^{j} \frac{1}{(j+1)^2}\sum\limits_{k=j}^{n}\binom{k}{j}\tag{5}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \sum_{j=0}^{k} (-1)^j\binom{k}{j}\frac{1}{(j+1)^2} \tag{6}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \frac{1}{k+1}\sum_{j=0}^{k} (-1)^j\binom{k+1}{j+1}\frac{1}{j+1}\tag{7}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \frac{1}{k+1}H_{k+1}\tag{8}\\ &= \frac{H_{n+1}^2+H_{n+1}^{(2)}}{2(n+1)^2}\tag{9}\end{align}$$
Justifications:
(1) Used $\displaystyle \binom{n+1}{k+1} = \frac{n+1}{k+1}\binom{n}{k}$ and applied change in order of summation.
(2) Used the identity: $\displaystyle \sum\limits_{k=j}^{n} (-1)^k\binom{n+1}{k+1} = \sum\limits_{k=j}^{n-1} (-1)^k\left(\binom{n}{k}+\binom{n}{k+1}\right) + (-1)^{n}\binom{n+1}{n+1} = (-1)^j\binom{n}{j}$ via telescoping summation.
(3) Used identity from (1).
(4) Used the identity: $\displaystyle \binom{n+1}{j+1} = \sum\limits_{k=j}^{n} \binom{k}{j}$
(5) Interchanged order of summation.
(6) Reused identity from (1).
(7) Used the identity: $\displaystyle \sum\limits_{j=0}^{n} (-1)^j\binom{n+1}{j+1}\frac{1}{j+1} = H_{n+1}$.
(8) Used the identity: $\displaystyle \sum\limits_{k=1}^{n} \frac{H_k}{k} = \frac{H_n^2 + H_n^{(2)}}{2}$.
More generally, for a sequence of numbers $\{a_k\}_{k\ge 0}$, we denote $\displaystyle b_n = \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} a_k$ and the binomial inversion formula tell us that $\displaystyle a_n = \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} b_k$. Denoting $\displaystyle s_n = \sum\limits_{k=0}^{n} a_k$ we have,
\begin{align*}
\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{s_k}{k+1} &= \frac{1}{n+1}\sum\limits_{k=0}^{n} (-1)^k \binom{n+1}{k+1} \sum\limits_{j=0}^{k} a_j \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} a_j \sum\limits_{k=j}^{n} (-1)^k \binom{n+1}{k+1} \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} a_j \sum\limits_{k=j}^{n} (-1)^k \left[\binom{n}{k} + \binom{n}{k+1}\right] \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} (-1)^j \binom{n}{j} a_j = \frac{b_n}{n+1}
\end{align*}
Let us denote, $\displaystyle c_{n,r} = \sum\limits_{j_1 = 1}^{n}\sum\limits_{j_2 = 1}^{j_1} \cdots \sum\limits_{j_r = 1}^{j_{r-1}} \frac{1}{j_1 \cdots j_r}$ then we have, $\displaystyle c_{n,r} = \sum\limits_{j=1}^{n} \frac{c_{j,r-1}}{j} = c_{n-1,r} + \frac{c_{n,r-1}}{n}$ and consequently,
\begin{align*}
\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} c_{k,r} &= \sum\limits_{k=1}^{n} (-1)^{k-1} \left[\binom{n-1}{k-1} + \binom{n-1}{k} \right] c_{k,r} \\&= \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n-1}{k-1} \left[c_{k-1,r} + \frac{c_{k,r-1}}{k}\right] + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k} c_{k,r} \\&= \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n-1}{k-1} c_{k-1,r} + \sum\limits_{k=1}^{n} (-1)^{k-1}\binom{n-1}{k-1}\frac{c_{k,r-1}}{k} + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k} c_{k,r} \\&= \sum\limits_{k=1}^{n-1} (-1)^{k} \binom{n-1}{k} c_{k,r} + \frac{1}{n}\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k}c_{k,r-1} + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k} c_{k,r} \\&= \frac{1}{n}\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k}c_{k,r-1}
\end{align*}
Therefore, by $r$ times application we have, $\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} c_{k,r} = \frac{1}{n^r}$ and from binomial inversion result we have, $\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k^r} = c_{n,r}$.
Then, using $\displaystyle a_k = \frac{1}{(k+1)^r}$ and denoting by classical notation $\displaystyle s_k = H_{k+1}^{(r)}$ we have,
\begin{align*}
\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{H_{k+1}^{(r)}}{k+1} &= \frac{1}{n+1}\sum\limits_{j=0}^{n} (-1)^j \binom{n}{j} \frac{1}{(j+1)^r} \\&= \frac{1}{(n+1)^2}\sum\limits_{j=1}^{n+1} (-1)^{j-1} \binom{n+1}{j} \frac{1}{j^{r-1}} = \frac{c_{n+1,r-1}}{(n+1)^2}
\end{align*}
Our particular case being $r=3$, we have, $$\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{H_{k+1}^{(3)}}{k+1} = \frac{c_{n+1,2}}{(n+1)^2} = \frac{H_{n+1}^2 + H_{n+1}^{(2)}}{2(n+1)^2}$$
The Binomial Inversion formula tells us:
$$\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}\frac{H_{k+1}^2+H_{k+1}^{(2)}}{2(k+1)^2} = \frac{H_{n+1}^{(3)}}{n+1}$$
Now at this point the Euler accelaration formula(or the Euler Series Transformation) tells us:
$$\sum\limits_{n=0}^{\infty} (-1)^{n}\frac{H_{n+1}^2+H_{n+1}^{(2)}}{2(n+1)^2} = \sum\limits_{n=0}^{\infty} \frac{H_{n+1}^{(3)}}{2^{n+1}(n+1)}$$
That is after reindexing the summation: $$\sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2} + \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = 2\sum\limits_{n=1}^{\infty}\frac{H_n^{(3)}}{2^nn}$$
The series in the RHS can be evaluated by observing that $\displaystyle \sum_{n=1}^{\infty}H_n^{(3)}x^n = \frac{\operatorname{Li}_3(x)}{1-x}$
Hence, $$\displaystyle \sum_{n=1}^{\infty}\frac{H_n^{(3)}}{n}x^{n} = \operatorname{Li}_4(x) -\frac{1}{2}\operatorname{Li}_2^2(x) - \log (1-x) \operatorname{Li}_3(x)$$
That is at $x = \dfrac{1}{2}$ we have: $$\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2} + \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = 2\sum\limits_{n=1}^{\infty} \frac{H_n^{(3)}}{2^{n}n} = 2\log 2 \operatorname{Li}_3\left(\frac{1}{2}\right) - \operatorname{Li}_2^2\left(\frac{1}{2}\right) + 2\operatorname{Li}_4\left(\frac{1}{2}\right)$$
Now it remains to deal with the alternating part: $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2}$ which was derived by R. Sitaramachandrarao in this paper.
$$\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = -\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$
An alternative approach of calculating the same is to observe that:
$$\begin{align} \sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^{(2)}}{n^2} &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4}+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{n-1}^{(2)}}{n^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\sum_{k=1}^{n-1}\frac1{k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=k+1}^{\infty}\frac{(-1)^{n-1}}{n^2k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n-1}}{(k+n)^2k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{k+n-1}\left(\frac1{k^2n(n+k)}-\frac1{kn(k+n)^2}\right)\\ &=\frac{7}{8}\zeta(4)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{k^2n(n+k)}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\ &=\frac{7}{8}\zeta(4)-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{k+n}\left(\frac{1}{kn^2(n+k)}+\frac{1}{k^2n(n+k)}\right)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\&= \frac{7}{8}\zeta(4)-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{n^2k^2}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\&=\frac{7}{8}\zeta(4)-\frac{1}{8}\zeta^2(2)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2} \\ &= \frac{21}{8}\zeta(4)-\frac{1}{8}\zeta^2(2)+2\sum_{k=1}^{\infty}\frac{(-1)^kH_k}{k^3}\end{align}$$
The later alternating series evaluates to: $$\sum_{k = 1}^{\infty} \frac{(-1)^k}{k^3}H_k = -\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3)$$
which can be computed by expressing it as a logarithmic integrals. User Chris'ssis had shown me in the past an interesting way of dealing with the integrals.
Update: We have the more general relation, $$\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{H_n}{n^3}x^n = -\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} -\frac{1}{2} \zeta(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(4)-\frac{1}{4}\operatorname{Li}_2^2(1-x)+\frac{1}{4}\zeta^2(2)+\operatorname{Li}_4(x)\\+\frac{1}{4}\log^2 x\log^2(1-x)+\frac{1}{2}\log x\log (1-x)\operatorname{Li}_2(1-x)+\zeta(3)\log x -\log x\operatorname{Li}_2(1-x)$$
where, we may write $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} = \zeta(2)\operatorname{Li}_2(1-x) + \operatorname{Li}_4(1-x) - \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2}(1-x)^n$.
I am too lazy to add a derivation for this. It's quite long and tedious for a blog post.
For those who are interested in similar logarithmic integrals can refer to Lewin's book Polylogarithms and associated functions pg. 310 formulae no. (10) to (12)). I don't have an elementary approach to this alternating Euler Sum.
We have seen how to deal with the Au-Yeung Series in three different ways in Closed form of Enrico Au-Yeung Series. However alternating versions (A) of Euler Sums often turn out to be trickier to deal with than the (H) counterpart.
Solution: We begin with the key identity:
$$\begin{align}\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}\frac{H_{k+1}^{(3)}}{k+1} &= \frac{1}{n+1}\sum_{k=0}^{n} (-1)^k \binom{n+1}{k+1}\sum\limits_{j=0}^{k}\frac{1}{(j+1)^3}\tag{1}\\ &= \frac{1}{n+1}\sum_{j=0}^{n}\frac{1}{(j+1)^3} \sum\limits_{k=j}^{n}(-1)^k \binom{n+1}{k+1}\tag{2}\\ &= \frac{1}{n+1}\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\frac{1}{(j+1)^3}\tag{3} \\ &= \frac{1}{(n+1)^2}\sum_{j=0}^{n}(-1)^{j} \binom{n+1}{j+1}\frac{1}{(j+1)^2} \tag{4}\\ &= \frac{1}{(n+1)^2}\sum_{j=0}^{n}(-1)^{j} \frac{1}{(j+1)^2}\sum\limits_{k=j}^{n}\binom{k}{j}\tag{5}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \sum_{j=0}^{k} (-1)^j\binom{k}{j}\frac{1}{(j+1)^2} \tag{6}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \frac{1}{k+1}\sum_{j=0}^{k} (-1)^j\binom{k+1}{j+1}\frac{1}{j+1}\tag{7}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \frac{1}{k+1}H_{k+1}\tag{8}\\ &= \frac{H_{n+1}^2+H_{n+1}^{(2)}}{2(n+1)^2}\tag{9}\end{align}$$
Justifications:
(1) Used $\displaystyle \binom{n+1}{k+1} = \frac{n+1}{k+1}\binom{n}{k}$ and applied change in order of summation.
(2) Used the identity: $\displaystyle \sum\limits_{k=j}^{n} (-1)^k\binom{n+1}{k+1} = \sum\limits_{k=j}^{n-1} (-1)^k\left(\binom{n}{k}+\binom{n}{k+1}\right) + (-1)^{n}\binom{n+1}{n+1} = (-1)^j\binom{n}{j}$ via telescoping summation.
(3) Used identity from (1).
(4) Used the identity: $\displaystyle \binom{n+1}{j+1} = \sum\limits_{k=j}^{n} \binom{k}{j}$
(5) Interchanged order of summation.
(6) Reused identity from (1).
(7) Used the identity: $\displaystyle \sum\limits_{j=0}^{n} (-1)^j\binom{n+1}{j+1}\frac{1}{j+1} = H_{n+1}$.
(8) Used the identity: $\displaystyle \sum\limits_{k=1}^{n} \frac{H_k}{k} = \frac{H_n^2 + H_n^{(2)}}{2}$.
More generally, for a sequence of numbers $\{a_k\}_{k\ge 0}$, we denote $\displaystyle b_n = \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} a_k$ and the binomial inversion formula tell us that $\displaystyle a_n = \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} b_k$. Denoting $\displaystyle s_n = \sum\limits_{k=0}^{n} a_k$ we have,
\begin{align*}
\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{s_k}{k+1} &= \frac{1}{n+1}\sum\limits_{k=0}^{n} (-1)^k \binom{n+1}{k+1} \sum\limits_{j=0}^{k} a_j \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} a_j \sum\limits_{k=j}^{n} (-1)^k \binom{n+1}{k+1} \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} a_j \sum\limits_{k=j}^{n} (-1)^k \left[\binom{n}{k} + \binom{n}{k+1}\right] \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} (-1)^j \binom{n}{j} a_j = \frac{b_n}{n+1}
\end{align*}
Let us denote, $\displaystyle c_{n,r} = \sum\limits_{j_1 = 1}^{n}\sum\limits_{j_2 = 1}^{j_1} \cdots \sum\limits_{j_r = 1}^{j_{r-1}} \frac{1}{j_1 \cdots j_r}$ then we have, $\displaystyle c_{n,r} = \sum\limits_{j=1}^{n} \frac{c_{j,r-1}}{j} = c_{n-1,r} + \frac{c_{n,r-1}}{n}$ and consequently,
\begin{align*}
\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} c_{k,r} &= \sum\limits_{k=1}^{n} (-1)^{k-1} \left[\binom{n-1}{k-1} + \binom{n-1}{k} \right] c_{k,r} \\&= \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n-1}{k-1} \left[c_{k-1,r} + \frac{c_{k,r-1}}{k}\right] + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k} c_{k,r} \\&= \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n-1}{k-1} c_{k-1,r} + \sum\limits_{k=1}^{n} (-1)^{k-1}\binom{n-1}{k-1}\frac{c_{k,r-1}}{k} + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k} c_{k,r} \\&= \sum\limits_{k=1}^{n-1} (-1)^{k} \binom{n-1}{k} c_{k,r} + \frac{1}{n}\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k}c_{k,r-1} + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k} c_{k,r} \\&= \frac{1}{n}\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k}c_{k,r-1}
\end{align*}
Therefore, by $r$ times application we have, $\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} c_{k,r} = \frac{1}{n^r}$ and from binomial inversion result we have, $\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k^r} = c_{n,r}$.
Then, using $\displaystyle a_k = \frac{1}{(k+1)^r}$ and denoting by classical notation $\displaystyle s_k = H_{k+1}^{(r)}$ we have,
\begin{align*}
\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{H_{k+1}^{(r)}}{k+1} &= \frac{1}{n+1}\sum\limits_{j=0}^{n} (-1)^j \binom{n}{j} \frac{1}{(j+1)^r} \\&= \frac{1}{(n+1)^2}\sum\limits_{j=1}^{n+1} (-1)^{j-1} \binom{n+1}{j} \frac{1}{j^{r-1}} = \frac{c_{n+1,r-1}}{(n+1)^2}
\end{align*}
Our particular case being $r=3$, we have, $$\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{H_{k+1}^{(3)}}{k+1} = \frac{c_{n+1,2}}{(n+1)^2} = \frac{H_{n+1}^2 + H_{n+1}^{(2)}}{2(n+1)^2}$$
The Binomial Inversion formula tells us:
$$\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}\frac{H_{k+1}^2+H_{k+1}^{(2)}}{2(k+1)^2} = \frac{H_{n+1}^{(3)}}{n+1}$$
Now at this point the Euler accelaration formula(or the Euler Series Transformation) tells us:
$$\sum\limits_{n=0}^{\infty} (-1)^{n}\frac{H_{n+1}^2+H_{n+1}^{(2)}}{2(n+1)^2} = \sum\limits_{n=0}^{\infty} \frac{H_{n+1}^{(3)}}{2^{n+1}(n+1)}$$
That is after reindexing the summation: $$\sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2} + \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = 2\sum\limits_{n=1}^{\infty}\frac{H_n^{(3)}}{2^nn}$$
The series in the RHS can be evaluated by observing that $\displaystyle \sum_{n=1}^{\infty}H_n^{(3)}x^n = \frac{\operatorname{Li}_3(x)}{1-x}$
Hence, $$\displaystyle \sum_{n=1}^{\infty}\frac{H_n^{(3)}}{n}x^{n} = \operatorname{Li}_4(x) -\frac{1}{2}\operatorname{Li}_2^2(x) - \log (1-x) \operatorname{Li}_3(x)$$
That is at $x = \dfrac{1}{2}$ we have: $$\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2} + \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = 2\sum\limits_{n=1}^{\infty} \frac{H_n^{(3)}}{2^{n}n} = 2\log 2 \operatorname{Li}_3\left(\frac{1}{2}\right) - \operatorname{Li}_2^2\left(\frac{1}{2}\right) + 2\operatorname{Li}_4\left(\frac{1}{2}\right)$$
Now it remains to deal with the alternating part: $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2}$ which was derived by R. Sitaramachandrarao in this paper.
$$\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = -\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$
An alternative approach of calculating the same is to observe that:
$$\begin{align} \sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^{(2)}}{n^2} &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4}+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{n-1}^{(2)}}{n^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\sum_{k=1}^{n-1}\frac1{k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=k+1}^{\infty}\frac{(-1)^{n-1}}{n^2k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n-1}}{(k+n)^2k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{k+n-1}\left(\frac1{k^2n(n+k)}-\frac1{kn(k+n)^2}\right)\\ &=\frac{7}{8}\zeta(4)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{k^2n(n+k)}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\ &=\frac{7}{8}\zeta(4)-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{k+n}\left(\frac{1}{kn^2(n+k)}+\frac{1}{k^2n(n+k)}\right)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\&= \frac{7}{8}\zeta(4)-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{n^2k^2}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\&=\frac{7}{8}\zeta(4)-\frac{1}{8}\zeta^2(2)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2} \\ &= \frac{21}{8}\zeta(4)-\frac{1}{8}\zeta^2(2)+2\sum_{k=1}^{\infty}\frac{(-1)^kH_k}{k^3}\end{align}$$
The later alternating series evaluates to: $$\sum_{k = 1}^{\infty} \frac{(-1)^k}{k^3}H_k = -\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3)$$
which can be computed by expressing it as a logarithmic integrals. User Chris'ssis had shown me in the past an interesting way of dealing with the integrals.
Update: We have the more general relation, $$\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{H_n}{n^3}x^n = -\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} -\frac{1}{2} \zeta(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(4)-\frac{1}{4}\operatorname{Li}_2^2(1-x)+\frac{1}{4}\zeta^2(2)+\operatorname{Li}_4(x)\\+\frac{1}{4}\log^2 x\log^2(1-x)+\frac{1}{2}\log x\log (1-x)\operatorname{Li}_2(1-x)+\zeta(3)\log x -\log x\operatorname{Li}_2(1-x)$$
where, we may write $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} = \zeta(2)\operatorname{Li}_2(1-x) + \operatorname{Li}_4(1-x) - \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2}(1-x)^n$.
I am too lazy to add a derivation for this. It's quite long and tedious for a blog post.
For those who are interested in similar logarithmic integrals can refer to Lewin's book Polylogarithms and associated functions pg. 310 formulae no. (10) to (12)). I don't have an elementary approach to this alternating Euler Sum.
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