Lemma 1 (see [8].)

• (i)

  Let (−1)ⁱdⁱ/dtⁱψ(t) > 0, t ∈ [m, ∞)(m ∈ N) and ψ⁽ⁱ⁾(∞) = 0 and (i = 0, 1, 2, 3), and let P_i(t), B_i(i ∈ N) denote, respectively, the Bernoulli functions and Bernoulli numbers of i-order. Then,

In particular, for q = 1 and B₂ = 1/6, we have the inequality:

For q = 2 and B₄ = −1/30, we have the inequality:

• (ii)

  If f(t)(>0) ∈ C³[m, ∞), f⁽ⁱ⁾(∞) = 0(i = 0, 1, 2, 3), then we have the following Euler-Maclaurin summation formula:

Lemma 2. For s ∈ (0, 6], s₂ ∈ (0, 2/β]∩(0, s), k_s(s₂) = B(s₂, s − s₂), we define the following weight coefficient:

Then, the following inequalities hold:

where $$

Proof. For fixed m ∈ N, we define a function ψ(m, t) as follows:

Using equality (Equation (9)), we have the following:

It is easy to observe that $$. In addition, integration by parts, it follows that

Further, for 0 < s₂ ≤ 2/β, 0 < β ≤ 1, s₂ < s ≤ 6, it is not difficult to verify that

By Equations (7)–(10), we obtain the following:

and then we get the following: where

It is easy to see that

where ψ(σ)(σ ∈ (0, (2/β)]) is defined by the following:

Further, we obtain that for β ∈ (0, 1], s ∈ (0, 6],

Thus, it follows that

Also, we find that for s₂ ∈ (0, (2/β)],

Hence, we have ℏ(m) > 0. Setting t = m^α/βu^1/β, it follows that

On the other hand, by Equation (9), we have the following:

Note that $$, and

For s₂ ∈ (0, 2/β]∩(0, s), 0 < s ≤ 6, by Equation (7), we find $$ and

Hence, we have the following:

and then we obtain the following: where $$ which satisfies the following:

Therefore, the two-sided inequalities in Equation (12) follow. This completes the proof of Lemma 2.

Lemma 3. Under the assumption (C1), we have the following reverse Hardy-Hilbert’s inequality:

Proof. By using the reverse Hӧlder’s inequality [9], we obtain the following:

Hence, by virtue of Equations (12) and (32), for s = λ, s_i = λ_i(i = 1, 2), we get inequality (Equation (33)). The proof of Lemma 3 is complete.

Lemma 4. Let t > 0, then we have the inequality:

Proof. In view of $$, applying the Abel partial summation formula, we find the following:

Since 1 − e^−t < t(t > 0), and for β ∈ (0, 1],

we acquire that which implies inequality (Equation (35)). Lemma 4 is proved.

Theorem 5. Under the assumption (C1), we have the following reverse Hardy-Hilbert’s inequality with one partial sum as the terms of double series:

In particular, for λ₁ + λ₂ = λ, we have the following:

and the following inequality:

Proof. In view of the fact that

by using inequality (Equation (35)), it follows that

Then by Equation (33), we deduce inequality (Equation (39)). The proof of Theorem 5 is complete.

Remark 6. For $$, in Equation (32), we have the following:

Theorem 7. Under the assumption (C1), if λ₁ + λ₂ = λ(∈(0, 5]), then for β = 1, λ₁ ∈ (0, 2/α]∩(0, λ), λ₂ ∈ (0, 1]∩(0, λ), the constant factor$$ in Equation (39) is the best possible.

Proof. We may divide two cases of λ₂ ∈ (0, 1)∩(0, λ) andλ₂ = 1(<λ) to prove that the constant factor 1/λα^1/qB(λ₁, λ₂) in Equation (41) (for β = 1) is the best possible.

Case (i) λ₂ ∈ (0, 1)∩(0, λ). For any 0 < ε < min{pλ₁, |q|(1 − λ₂)}, we set the following:

Since 0 < λ₂ − ε/q < 1, by the decreasingness property of series, we have the following:

If there exists a constant M, M ≥ 1/λα^1/qB(λ₁, λ₂), such that Equation (41) for (β = 1) is valid when 1/λα^1/qB(λ₁, λ₂) is replaced by M; then, in particular, by substitution of $$, and $$ in Equation (41) for (β = 1), we have the following:

By using inequality (Equation (47)) and the decreasingness property of series, we obtain the following:

By utilizing inequality (Equation (44)) and setting $$, we find the following:

Then, we have

Letting ε⟶0⁺ along with the continuity of the gamma function, we deduce that 1/λα^1/qB(λ₁, λ₂) ≥ M. Hence, M = 1/λα^1/qB(λ₁, λ₂) is the best possible constant factor of Equation (41) (for β = 1).

Case (ii) λ₂ = 1(<λ). For 0 < ε < 1, replacing λ by λ − ε in Equation (41) and taking λ₁ = λ − 1, λ₂ = 1 − ε, and β = 1, we obtain the following:

where $$.

If λ₁ = 1, then we have the following:

It follows that $$, and similarly, $$.

If there exists a constant factor M, M ≥ 1/α^1/qλB(λ − 1, 1) = 1/α^1/qλ(λ − 1), such that (Equation (41)) for (β = 1, λ₂ = 1, λ₁ = λ − 1) is valid when we replace 1/α^1/qλ(λ − 1) by M, namely,

then, by applying Fatou’s lemma [10] and Equation (53), we deduce that

By the property of limitation, there exists a constant δ₀ ∈ (0, 1), such that for any δ ∈ (0, δ₀),

namely,

By Case (i), if the constant factor 1/α^1/q(λ − δ)B(λ − 1, 1 − δ) (Equation (51)) ε = δ is the best possible, we have 1/α^1/q(λ − δ)B(λ − 1, 1 − δ) ≥ M. Then, for δ⟶0⁺, we have the following:

which implies that M = 1/α^1/qλ(λ − 1) is the best possible factor of Equation (41) (for (β = λ₁ = 1, λ₂ = λ − 1)). This completes the proof of Theorem 7.

Theorem 8. Under the assumption (C1), if the constant factor $$ in Equation (39) is the best possible, then for

we have λ₁ + λ₂ = λ..

Proof. For $$, we have the following:

If λ − λ₁ − λ₂ ∈ (−λ₁p, (2/α − λ₁)p](⊃{0}), then $$; if

then $$.

It is easy to observe that $$, and then, by Equation (58), it follows that $$, $$, and $$. By Equation (41), we have the following:

If the constant factor $$ in Equation (39) is the best possible, then by Equation (61), we have the inequality:

that is

By the reverse Hӧlder’s inequality [9], we obtain the following:

In light of Equation (63), we deduce that Equation (64) keeps the form of equality. Thus, there exist constants A and B, such that they are not both zero and they satisfy $$ in R₊ (see [9]). Assuming that A ≠ 0, we have $$ a.e. in R₊, and hence, λ − λ₂ − λ₁ = 0, namely, λ₁ + λ₂ = λ. Theorem 8 is proved.

Theorem 9. Under the assumption (C1), we have the following inequality which is equivalent to Equation (39):

In particular, for λ₁ + λ₂ = λ, we have the following:

and the following inequality which is equivalent to Equation (41):

Proof. Suppose that Equation (65) is valid. By the reverse Hӧlder’s inequality [9], we have the following:

Then by Equation (65), we obtain Equation (39). On the other hand, assuming that Equation (39) is valid, we set the following:

Then, it follows that $$.

If J₁ = ∞, then Equation (65) is naturally valid; if J₁ = 0, then it is impossible that makes Equation (65) valid, namely, J₁ > 0. Suppose that 0 < J₁ < ∞. By Equation (39), we have the following:

namely, Equation (65) follows, which is equivalent to Equation (39). Theorem 9 is proved.

Remark 10. By the same way as above, under the assumption (C1), if p < 0, 0 < q < 1, we can obtain the following equivalent inequalities:

Theorem 11. Under the assumption (C1), if λ₁ + λ₂ = λ(∈(0, 5]), then for β = 1, λ₁ ∈ (0, 2/α]∩(0, λ), and λ₂ ∈ (0, 1]∩(0, λ), the constant factor $$ in Equation (65) is the best possible. On the other hand, if the same constant factor in Equation (65) is the best possible, then for

we have λ₁ + λ₂ = λ..

Proof. If λ₁ + λ₂ = λ, then by Theorem 7, the constant factor 1/α^1/qλB(λ₁, λ₂) in Equation (41) (for β = 1) is the best possible. Then by Equation (68) (for λ₁ + λ₂ = λ, β = 1), the constant factor in Equation (65) is still the best possible.

On the other hand, if the same constant factor, $$, in Equation (65) is the best possible, then by the equivalency of Equations (65) and (39), in view of I = J^q (in the proof of Theorem 9), it follows that the same constant factor in Equation (39) is the best possible. By using Theorem 7, we obtain λ₁ + λ₂ = λ. The proof of Theorem 11 is complete.

Remark 12. (i) Choosing β = 1, λ = 1, λ₁ = 1/r, λ₂ = 1/s(r > 1, 1/r + 1/s = 1) in Equations (41) and (67), respectively, we acquire the following inequalities with the best possible constant factor π/α^1/qsin(π/r):

In particular, for r = s = 2, we deduce from Equation (68) and Equation (71) that

(ii) Putting β = 1, λ = 3, λ₁ = 2, λ₂ = 1 in Equations (41) and (67), respectively, we obtain the following inequalities with the best possible constant factor 1/6α^1/q