The sum: $\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}$

Since $[(H_k)^2+ H_k^{(2)}]=2\sum_{j=1}^k\frac{H_j}{j}$ it follows that, $$\begin{align} \sum_{k=1}^{n}(-1)^{k-1}[(H_k)^2+ H_k^{(2)}] \binom{n}{k}&= 2\sum_{k=1}^{n}(-1)^{k-1}{n \choose k}\sum_{j=1}^k\frac{H_j}{j}\\ &=2\sum_{j=1}^n\frac{H_j}{j}\sum_{k=j}^{n}(-1)^{k-1}\binom{n}{k} \\ &=2\sum_{j=1}^n\frac{H_j}{j} (-1)^{j-1}\binom{n-1}{j-1}\\ &=\frac{2}{n}\sum_{j=1}^n (-1)^{j-1}\binom{n}{j}H_j=\frac{2}{n^2}. \end{align}$$ The known identities we used can be found in this paper: "Some identities involving Harmonic Numbers" by J. Spiess.

Recalling that $$\int_{0}^{1}x^{a-1}\log^{m}\left(x\right)\log\left(1-x\right)^{n}dx=\frac{\partial^{m+n}}{\partial a^{m}\partial b^{n}}B\left(a,b\right)\mid_{b=1}$$ where $B\left(a,b\right)$ is the Beta function and since $$H_{\alpha}^{\left(m\right)}=\frac{\left(-1\right)^{m-1}}{\left(m-1\right)!}\left(\psi^{\left(m-1\right)}\left(\alpha+1\right)-\psi^{\left(m-1\right)}\left(1\right)\right)$$ where $\alpha\notin\mathbb{N}^{-}$, $m\geq2$ and where $\psi^{\left(m-1\right)}\left(x\right)$ is the polygamma function, we can prove that $$\int_{0}^{1}x^{k-1}\log\left(1-x\right)^{2}dx=\frac{H_{k}^{2}+H_{k}^{\left(2\right)}}{k}$$ hence $$\sum_{k=1}^{n}\dbinom{n}{k}\left(-1\right)^{k-1}\left(H_{k}^{2}+H_{k}^{\left(2\right)}\right)=\int_{0}^{1}\sum_{k=1}^{n}\dbinom{n}{k}\left(-1\right)^{k-1}kx^{k-1}\log\left(1-x\right)^{2}dx$$ $$=n\int_{0}^{1}\left(1-x\right)^{n-1}\log\left(1-x\right)^{2}dx=\color{red}{\frac{2}{n^{2}}}.$$

Let $$S_n=\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}] {n \choose k}.$$ Notice that $$H_k^2+H_k^{(2)} =\left(\sum_{j=1}^{k} \frac{1}{j}\right)^2+\sum_{j=1}^{k} \frac{1}{j^2}=2 \sum \sum_{1 \le i \le j \le k} \frac{1}{ij}$$ So $S_n$ can be re-written as $$S_n=2\sum_{i=1}^{n} \frac{1}{i} \sum _{j=i}^{n} \frac{1}{j} \sum_{k=j}^{n} (-1)^{k-1} {n \choose k}=2 \sum_{i=1}^{n} \frac{1}{i} \sum _{j=i}^{n} \frac{1}{j} \sum_{k=0}^{j-1} (-1)^{k} {n \choose k}= 2 \sum_{i=1}^{n} \frac{1}{i} \sum _{j=i}^{n} \frac{1}{j} (-1)^j {n-1 \choose j-1}.$$ Here we have used $$\sum_{k=j}^{n} (-1)^{k-1} {n \choose k}= 0-\sum_{k=0}^{j-1} (-1)^{k-1} {n \choose k}, \sum_{k=0}^m (-1)^k {n \choose k} =(-1)^m {n-1 \choose m}~~~(1)$$ Next we use $${n-1 \choose m-1}=\frac{m}{n} {n \choose n}~~~~(2)$$ twice to get $$S_n=2\sum_{i=1}^{n} \frac{1}{i} \sum_{j=i}^{n} \frac{1}{j} (-1)^{j}~ \frac{j} {n} {n \choose j}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} \sum_{j=i}^{n} (-1)^j {n \choose j}.$$ Again using (1) and (2), we get $$ \Rightarrow S_n=-\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} \sum_{j=0}^{i-1} (-1)^j {n \choose j}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} (-1)^i {n-1 \choose i-1}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} (-1)^i \frac{i}{n}{n \choose i}$$ $$\Rightarrow S_n=\frac{2}{n^2}.$$