An Extension of a Congruence by Tauraso

For a positive integer $n$ let $H_n={\sum }_{k=1}^{n}1\mathrm{/}n$ be the $n$th harmonic number. In this note we prove that for any prime $p\ge 7$, $$\sum _{k=1}^{p-1}\frac{H_k}{k^2}\equiv \sum _{k=1}^{p-1}\frac{H_k^2}{k}\equiv \frac{3}{2p}\sum _{k=1}^{p-1}\frac{1}{k^2}(\mathrm{m}\mathrm{o}\mathrm{d}{p}^2)\mathrm{.}$$ Notice that the first part of this congruence is recently proposed by R. Tauraso as a problem in Amer. Math. Monthly. In our elementary proof of the second part of the above congruence we use certain classical congruences modulo a prime and the square of a prime, some congruences involving harmonic numbers and a combinatorial identity due to V. Hern'{a}ndez.