An Extension of a Congruence by Tauraso

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.


Introduction and Main Results
Given positive integers and , the harmonic numbers of order are those rational numbers de�ned as = = . (1) For simplicity, we will denote by the th harmonic number (we assume in addition that 0 = 0 = 0). Usually, here as always in the sequel, we consider the congruence relation modulo a prime extended to the ring of rational numbers with denominators not divisible by . For such fractions we put / ≡ / mod if and only if ≡ mod , and the residue class of / is the residue class of where is the inverse of modulo . By a problem proposed by Tauraso in [1] and recently solved by Tyler [2], for any prime 7, = 2 ≡ = 2 mod 2 .
Further, Tauraso [ Tauraso's proof of (4) is based on an identity due to Hernández [4] (see Lemma 8) and the congruence for triple harmonic sum modulo a prime due to Zhao [5] (see (64) of Remarks in Section 2). In this paper, we give an elementary proof of (4) and its extension as follows.
Recall that Sun in [6] established basic congruences modulo a prime for several sums of terms involving harmonic numbers. In particular, Sun established ∑ = mod 4 for = 2 . Further generalizations of these congruences are recently obtained by Tauraso in [7].
Recall that the Bernoulli numbers are de�ned by the generating function In particular, one has Remark 3. Notice that the second congruence of (8) was obtained by Sun and Tauraso [9, the congruence (5.4)] by using a standard technique expressing sum of powers in terms of Bernoulli numbers. Our proof of the second part of the congruence (5) given in the next section is entirely elementary and it is combinatorial in spirit. It is based on certain classical congruences modulo a prime and the square of a prime, two simple congruences given by Sun [6], and two particular cases of a combinatorial identity due to Hernández [4].

Proof of Theorem 1
e following congruences given by Sun in his recent paper [6] are needed in the proof of eorem 1.
e above identity and (13) and (11) of Lemma 5 with = 4 yield On the other hand, since by (9) from Lemma 4, ≡ (mod for each , then Taking (22) into (21) gives which proves (14). Proof of the congruence (15) is completely analogous to the previous proof using the fact that, by Lemma 5, 3 ≡ 0 (mod and therefore, for each Finally (cf. [2]), from the identity immediately follows that Inserting in the right hand side of the identity (26) the congruences ≡ 3 ≡ 0 (mod given in Lemma 5, we immediately obtain (16). is completes the proof.

Lemma 7.
Let be a prime. en Proof. Since + / , for every , we get Using particular congruences given in Lemma 5 with and 4, we �nd that Substituting the congruences (29), (13) of Lemma 6, and (11) with 4 of Lemma 5 into (28), we obtain e right hand side of (30) can be expressed as Taking (15) of Lemma 6 into (31) and comparing this with (30), we immediately obtain (27).
Further, for the proof of eorem 1 we will need two particular cases of the following identity due to Hernández [4].
Inserting in the right hand side of the above identity the congruences ≡ 0( mod 2 ) and However, the determination of ∑ = ( 2 / ) (mod 3 ) seems to be a difficult problem.