An Extension of a Congruence by Tauraso
  For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/n$ be the $n$th harmonic number. In this note we prove that for any prime $p\ge 7$,   $$ \sum_{k=1}^{p-1}\frac{H_k}{k^2}\equiv \sum_{k=1}^{p-1}\frac{H_k^2}{k} \equiv\frac{3}{2p}\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}.   $$ 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. 
Romeo Mestrovic
math.NT