The sum: $\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}$ This attractive identity that
$$\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}~~~(*)$$
 emerged while doing numerics at Mathematica with harmonic numbers, binomial coefficients
 and sums involving them.
Here $$H_k=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k},~~ H_k^{(2)}=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{k^2}. $$
The question is: How to prove it $(*)$ by hand?
 A: 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.
A: 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 $$\sum_{k=1}^{n}\dbinom{n}{k}\left(-1\right)^{k-1}\left(H_{k}^{2}+H_{k}^{\left(2\right)}\right)=\int_{0}^{1}\sum_{k=1}^{n}\dbinom{n}{k}\left(-1\right)^{k-1}kx^{k-1}\log\left(1-x\right)^{2}dx$$ $$=n\int_{0}^{1}\left(1-x\right)^{n-1}\log\left(1-x\right)^{2}dx=\color{red}{\frac{2}{n^{2}}}.$$
A: 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
$$ \Rightarrow S_n=-\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} \sum_{j=0}^{i-1} (-1)^j {n \choose j}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} (-1)^i  {n-1 \choose i-1}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} (-1)^i \frac{i}{n}{n \choose i}$$ $$\Rightarrow S_n=\frac{2}{n^2}.$$
A: The answer for your question is a special case of the following reversal, using $~a_k, b_k\in\mathbb{C}~$:  

$$b_n=\sum_{k=0}^n (-1)^k{\binom n k}a_k \enspace\Leftrightarrow\enspace a_n=\sum_{k=0}^n (-1)^k{\binom n k}b_k$$

With $~a_0=b_0:=0~$ and $~\displaystyle a_k:=(H_k)^2+H_k^{(2)}~,\enspace b_k:=-\frac{2}{k^2}~$ we get:

$$(H_n)^2+H_n^{(2)} = 2\sum\limits_{k=1}^n\frac{(-1)^{k-1}}{k^2}{\binom n k}$$

Using $~\displaystyle ((H_{n+1})^2+H_{n+1}^{(2)})- ((H_n)^2+H_n^{(2)}) = \frac{2}{n+1}H_{n+1}~$ and knowing, 
that $~\displaystyle H_n=\sum\limits_{k=1}^n\frac{(-1)^{k-1}}{k}{\binom n k}~$, it’s very easy to prove the formula above by induction, 
which I leave for you as an exercise.
