A prime $p$ is said to be a Wolstenholme prime if it satisfies the congruence ${2p-1\choose p-1} \equiv 1 \pmod {p^4}$. For such a prime $p$, we establish an expression for ${2p-1\choose p-1}\pmod {p^8}$ given in terms of the sums $R_i:=\sum _{k=1}^{p-1}1/k^i$ ($i=1,2,3,4,5,6)$. Further, the expression in this congruence is reduced in terms of the sums $R_i$ ($i=1,3,4,5$). Using this congruence, we prove that for any Wolstenholme prime $p$ we have $$ \left ({2p-1\atop p-1}\right ) \equiv 1 -2p \sum _{k=1}^{p-1}\frac {1}{k} -2p^2\sum _{k=1}^{p-1}\frac {1}{k^2}\pmod {p^7}. $$ Moreover, using a recent result of the author, we prove that a prime $p$ satisfying the above congruence must necessarily be a Wolstenholme prime. Furthermore, applying a technique of Helou and Terjanian, the above congruence is given as an expression involving the Bernoulli numbers.
Let $p>3$ be a prime, and let $q_p(2)=(2^{p-1}-1)/p$ be the Fermat quotient of $p$ to base $2$. In this note we prove that $$ \sum _{k=1}^{p-1} \frac {1}{k\cdot 2^k} \equiv q_p(2)-\frac {pq_p(2)^2}{2}+ \frac {p^2 q_p(2)^3}{3} -\frac {7}{48} p^2 B_{p-3}\pmod {p^3}, $$ which is a generalization of a congruence due to Z. H. Sun. Our proof is based on certain combinatorial identities and congruences for some alternating harmonic sums. Combining the above congruence with two congruences by Z. H. Sun, we show that $$ q_p(2)^3 \equiv -3\sum _{k=1}^{p-1} \frac {2^k}{k^3}+ \frac {7}{16} \sum _{k=1}^{(p-1)/2} \frac {1}{k^3} \pmod {p}, $$ which is just a result established by K. Dilcher and L. Skula. As another application, we obtain a congruence for the sum $\sum _{k=1}^{p-1}1/(k^2\cdot 2^k)$ modulo $p^2$ that also generalizes a related Sun's congruence modulo $p$.