<s>[INST] 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?
 [/INST] 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
$$ \Rightarrow S_n=-\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} \sum_{j=0}^{i-1} (-1)^j {n \choose j}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} (-1)^i  {n-1 \choose i-1}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} (-1)^i \frac{i}{n}{n \choose i}$$ $$\Rightarrow S_n=\frac{2}{n^2}.$$
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