  For $k=1,2,\ldots$ let $H_k$ denote the harmonic number $\sum_{j=1}^k 1/j$.
In this paper we establish some new congruences involving harmonic numbers. For
example, we show that for any prime $p>3$ we have
$$\sum_{k=1}^{p-1}\frac{H_k}{k2^k}\equiv\frac7{24}pB_{p-3}\pmod{p^2},\ \
\sum_{k=1}^{p-1}\frac{H_{k,2}}{k2^k}\equiv-\frac 38B_{p-3}\pmod{p},$$ and
$$\sum_{k=1}^{p-1}\frac{H_{k,2n}^2}{k^{2n}}\equiv\frac{\binom{6n+1}{2n-1}+n}{6n+1}pB_{p-1-6n}\pmod{p^2}$$
for any positive integer $n<(p-1)/6$, where $B_0,B_1,B_2,\ldots$ are Bernoulli
numbers, and $H_{k,m}:=\sum_{j=1}^k 1/j^m$.
