TITLE: The sum: $\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}$
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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.