  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.
