This is complementary to the [note](https://www.math.temple.edu/~tewodros/Harmonic_Sumsvnew.pdf)
 by Amdeberhan and Tauraso. In that note, Amdeberhan and Tauraso [Equations (12) and (15)] showed that for primes $p\ge 5$,
\begin{align}
&\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}H_k\equiv
\frac{1}{2}q_p(2)^2+(-1)^{\frac{p-1}{2}}E_{p-3}\pmod{p},\tag{1}\\
&\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}H_{2k}\equiv
\frac{1}{4}q_p(2)^2\pmod{p}.\tag{2}
\end{align}
Here $H_n$ denotes the $n$-th harmonic number, $E_n$ denotes the $n$-th Euler number, and
the Fermat quotient of an integer $a$ with respect to an odd prime $p$ is given by
$q_p(a)=(a^{p-1}-1)/p$. 

**We shall give alternative proofs of (1) and (2) by use of combinatorial identities.**

**Proof of (1).**
For $0\le k\le p-1$, we have
\begin{align*}
{p-1\choose k}=\frac{(p-1)(p-2)\cdots(p-k)}{k!}\equiv (-1)^k+p(-1)^{k-1}H_k\pmod{p^2},
\end{align*}
and so
\begin{align}
(-1)^kH_k\equiv \frac{1}{p}\left((-1)^k-{p-1\choose k}\right)\pmod{p}.\tag{3}
\end{align}
It follows that
\begin{align}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^{k-1}}{k}H_{k-1}
&\equiv -\frac{1}{p}\left(\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}
+\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k}{p-1\choose k-1}\right)\pmod{p}.\tag{4}
\end{align}

Note that
\begin{align}
\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k}{p-1\choose k-1}
=\frac{1}{p}\sum_{k=1}^{\frac{p-1}{2}}{p\choose k}
=\frac{1}{2p}\left(\sum_{k=0}^{p-1}{p\choose k}-2\right)=q_p(2).\tag{5}
\end{align}
Next, we shall prove that
\begin{align}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\equiv -q_p(2)+\frac{p}{2}q_p(2)^2-p(-1)^{\frac{p-1}{2}}E_{p-3}\pmod{p^2}.\tag{6}
\end{align}
By Lehmer's congruence [Ann. Math. 39 (1938), 350--360, (45)],
we deduce that
\begin{align}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}=
\sum_{k=1}^{\lfloor p/4\rfloor}\frac{1}{k}-\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k}
\equiv \sum_{k=1}^{\lfloor p/4\rfloor}\frac{1}{k}+2q_p(2)-pq_p(2)^2\pmod{p^2},\tag{7}
\end{align}
where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to a real number $x$.
Another congruence due to Lehmer [Ann. Math. 39 (1938), 350--360, (43)] says
\begin{align}
\sum_{k=1}^{\lfloor p/4\rfloor}\frac{1}{p-4k}\equiv \frac{3}{4}q_p(2)-\frac{3}{8}pq_p(2)^2\pmod{p^2}.\tag{8}
\end{align}
For $0\le k \le\lfloor\frac{p}{4}\rfloor$, we have
\begin{align*}
\frac{1}{p-4k}\equiv -\frac{1}{4k}-\frac{p}{16k^2}\pmod{p^2}.
\end{align*}
Substituting the above into (8) gives
\begin{align}
\sum_{k=1}^{\lfloor p/4\rfloor}\frac{1}{k}\equiv
-3q_p(2)+\frac{3}{2}pq_p(2)^2-\frac{p}{4}\sum_{k=1}^{\lfloor p/4 \rfloor}\frac{1}{k^2}\pmod{p^2}.\tag{9}
\end{align}
By [Ann. Math. 39 (1938), 350--360, (48)], we have
\begin{align}
\sum_{k=1}^{\lfloor p/4\rfloor}\frac{1}{k^2}\equiv 4(-1)^{\frac{p-1}{2}}E_{p-3}\pmod{p}.\tag{10}
\end{align}
Combining (7), (9) and (10), we are led to (6).

Applying (5) and (6) to the right-hand side of (4), we obtain
\begin{align}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^{k-1}}{k}H_{k-1}
\equiv -\frac{1}{2}q_p(2)^2+(-1)^{\frac{p-1}{2}}E_{p-3}\pmod{p}.\tag{11}
\end{align}
Sun [Sci. China Math. 54 (2011), 2509--2535, Lemma 2.4] showed that
\begin{align}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k^2}\equiv (-1)^{\frac{p-1}{2}}2E_{p-3}\pmod{p}.\tag{12}
\end{align}
From the above and (11), we deduce that
\begin{align*}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^{k}}{k}H_{k}
&=\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^{k}}{k^2}-\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^{k-1}}{k}H_{k-1}\\
&\equiv \frac{1}{2}q_p(2)^2+(-1)^{\frac{p-1}{2}}E_{p-3}\pmod{p}.
\end{align*}
The desired result is reached.

**Proof of (2).**
By (3), we obtain that for $0\le k\le \frac{p-1}{2}$,
\begin{align*}
H_{2k-1}\equiv \frac{1}{p}\left(1+{p-1\choose 2k-1}\right)\pmod{p}.
\end{align*}
It follows that
\begin{align}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}H_{2k-1}
\equiv \frac{1}{p}\left(\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}
+\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}{p-1\choose 2k-1}\right)\pmod{p}.\tag{13}
\end{align}
Letting $n=\frac{p-1}{2}$ in the following identity:
\begin{align*}
\sum_{k=0}^n(-1)^k{2n+1\choose 2k}=(-1)^{\frac{n(n+1)}{2}}2^n,
\end{align*}
which can be easily proved by Zeilberger's algorithm, we find that
\begin{align}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}{p-1\choose 2k-1}
=\frac{2}{p}\left(\sum_{k=0}^{\frac{p-1}{2}}(-1)^k{p\choose 2k}-1\right)
=\frac{2}{p}\left((-1)^{\frac{p^2-1}{8}}2^{\frac{p-1}{2}}-1\right).\tag{14}
\end{align}

Next, we show that
\begin{align}
\frac{2}{p}\left((-1)^{\frac{p^2-1}{8}}2^{\frac{p-1}{2}}-1\right)
\equiv q_p(2)-\frac{p}{4}q_p(2)^2\pmod{p^2},\tag{15}
\end{align}
which is equivalent to
\begin{align}
8(-1)^{\frac{p^2-1}{8}}2^{\frac{p-1}{2}}+\left(2^{\frac{p-1}{2}}\right)^4
-6\left(2^{\frac{p-1}{2}}\right)^2-3\equiv 0 \pmod{p^3}.\tag{16}
\end{align}
Note that
\begin{align*}
(-1)^{\frac{p^2-1}{8}}=\left(\frac{2}{p}\right),
\end{align*}
where $\left(\frac{\cdot}{p}\right)$ denotes the Legendre symbol.
If $\left(\frac{2}{p}\right)=1$, then the left-hand side of (16) equals
\begin{align*}
\left(2^{\frac{p-1}{2}}+3\right)\left(2^{\frac{p-1}{2}}-1\right)^3\equiv 0 \pmod{p^3}.
\end{align*}
If $\left(\frac{2}{p}\right)=-1$, we find that the left-hand side of (16) equals
\begin{align*}
\left(2^{\frac{p-1}{2}}-3\right)\left(2^{\frac{p-1}{2}}+1\right)^3\equiv 0 \pmod{p^3}.
\end{align*}
Now we conclude the proof of (16).

Combining (6) and (13)--(15), we obtain
\begin{align*}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}H_{2k-1}
\equiv \frac{1}{4}q_p(2)^2-(-1)^{\frac{p-1}{2}}E_{p-3}\pmod{p}.
\end{align*}
It follows from the above and (12) that
\begin{align*}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}H_{2k}
=\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}H_{2k-1}+\frac{1}{2}
\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k^2}\equiv \frac{1}{4}q_p(2)^2\pmod{p},
\end{align*}
which completes the proof of (2).