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Numerical calculation suggests that for prime $p\ge 5$, \begin{align*} \sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}\equiv 0\pmod{p}. \end{align*}

Question. How can we prove this congruence?

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It turned out the OP's question required a good amount of work and a joint effort (with Roberto Tauraso). However, "the margin here is too small to contain the proof", so to say.

Instead, readers can find the resolution at this link. In addition, we prove that \begin{align*} \sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}=-\frac58\zeta(2). \end{align*}

Comments and suggestions are welcome.

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    $\begingroup$ Little comment: $p'$ is used before it is defined. Congrats on solving this! (I spent some hours to no avail.) $\endgroup$ – darij grinberg Jun 21 '17 at 22:05
  • $\begingroup$ @ Professor Amdeberhan: Thank you very much for your great efforts to solve this problem. Thanks also go to Professor Tauraso. $\endgroup$ – Ji-Cai Liu Jun 26 '17 at 10:54
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This is complementary to the note by Amdeberhan and Tauraso. In order to solve this problem, Amdeberhan and Tauraso [Equations (12) and (15)] established the following two important results: \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} for primes $p\ge 5$. 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).

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