New bounds for the exponential function with cotangent

In this paper, new bounds for the exponential function with cotangent are found by using the recurrence relation between coefficients in the expansion of power series of the function ln(1−2x2/15−px6)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\ln (1-2x^{2}/15-px^{6})$\end{document} and a new criterion for the monotonicity of the quotient of two power series.

Throughout the full text, we suppose that where p 0 is the unique zero of the function h(p) = ln 1 -π 2 30 -π 6 64 p -8(45π 4 p + 32) 15π 6 p + 32π 2 -960 (1.4) on (0, p 2 ). Now considering the asymptotic expansion of (e x cot x-1 ) 2/5 , we have e x cot x-1 2/5 = 1 -2 15 x 2 -4 70,875 x 6 + 2 1,063,125 It is interesting that the power series above has not item of x 4 , which also remind us to establish a more accurate estimate for exp(x cot x -1). The first aim of this paper is to determine the best parameters p and q such that the double inequality holds for all x ∈ (0, π/2). The main conclusions of this paper are proved by the recursive method and a new criterion for the monotonicity of the quotient of two power series.
The following result is a theorem on the recurrence relation of coefficients in the series expansion of the function ln(1 -2x 2 /15px 6 ).
Our main results are contained in the following theorems.
is strictly increasing on (0, π/2), and therefore the double inequality p .
As a consequence of Theorem 2, we immediately get the following. for all x ∈ (0, π/2).
The second aim of this paper is to refine some known results presented in [19], we shall state it carefully in the fifth section.

Lemmas
In this paper, we will use some methods, such as the monotone form of l'Hospital's rule, an important criterion for the monotonicity of the quotient of two power series, and the latest promotion of the latter.
Now, we will introduce a useful auxiliary function H f ,g . For -∞ ≤ a < b ≤ ∞, let f and g be differentiable on (a, b) and g = 0 on (a, b). Then the function H f ,g is defined by The function H f ,g has some good properties [25, Property 1] and plays an important role in the proof of a monotonicity criterion for the quotient of power series (see [26]).

Lemma 3 ([26, Theorem 1]) Let A(t)
= ∞ k=0 a k t k and B(t) = ∞ k=0 b k t k be two real power series converging on (-r, r) and b k > 0 for all k. Suppose that for certain m ∈ N, the nonconstant sequence {a k /b k } is increasing (resp. decreasing) for 0 ≤ k ≤ m and decreasing (resp. increasing) for k ≥ m. Then the function A/B is strictly increasing (resp. decreasing) on (0, r) if and only if H A,B (r -) ≥ (resp. ≤) 0. Moreover, if H A,B (r -) < (resp. >) 0, then there exists t 0 ∈ (0, r) such that the function A/B is strictly increasing (resp. decreasing) on (0, t 0 ) and strictly decreasing (resp. increasing) on (t 0 , r).
Comparing coefficients gives the recurrence formulas (1.7) and (1.6). From the second equality of (3.1) we easily find that a n > 0 for all n ≥ 1, which completes the proof.

Proofs of Theorems 2 and 3
Proof of Theorem 2 Using the expansion the function f /g can be expressed as (2n)! |B 2n |x 2n := ∞ n=1 a n t n ∞ n=1 b n t n by Theorem 1, where x 2 = t. We now observe the monotonicity of the sequence {a n /b n } n≥1 . Since b n > 0 for all n ≥ 1, it suffices to determine the sign of c n := a n+1 -(b n+1 /b n )a n . Direct computations yield We claim that c n > 0 for n ≥ 3. In fact, by means of the recurrence formula (1.7), we have c n = a n+1 -b n+1 b n a n = a n+1 -2 (n + 1)(2n + 1) |B 2n+2 | |B 2n | a n = 2 n + 1 n 15 -1 2n + 1 |B 2n+2 | |B 2n | a n + p n -2 n + 1 a n-2 .
Clearly, if we prove for n ≥ 3, then it follows that c n > 0. Using the right hand side of (2. for n ≥ 4. This together with d 3 = 0 yields d n ≥ 0 for n ≥ 3. (i) If p 2 ≤ p < p 3 , then c n = a n+1 -(b n+1 /b n )a n > 0 for n ≥ 1, that is, the sequence {a n /b n } n≥1 is strictly increasing, so is f /g on (0, π/2) by Lemma 2. Therefore, we conclude that which implies (1.8).

Proof of Theorems 3 Let
we find that the function H(p, x) is decreasing with respect to p on (0, p 3 ). Then by the left hand side of (1.8) and by (1.10) we can complete the proof of Theorem 3.

Consequences and remarks
Remark 1 One can obtain the double inequality (1.12) using the key theorem of Wu and Debnath [28].
Let p → 0 + in (iii) of Theorem 2. Then we have the following.
Proof The necessity follows from To prove the sufficiency, we note that where f 1 is positive and decreasing on (0, π/2) by Corollary 1, it thus suffices to prove the function f 2 is positive and decreasing on (0, π/2). A simple computation gives which indicates that f 2 is strictly decreasing on (0, π/2) by Lemma 1. Meanwhile f 2 (x) is obviously positive for p ∈ (0, 2/15], which proves the sufficiency. Inequalities (5.3) follow from the decreasing property of the function F p (x). The proof is finished.
Remark 3 We claim that the lower bound in the double inequality (5.3) is strictly increasing with respect to the parameter p. In fact, put with h 1 (0 + ) = h 2 (0 + ) = 0, then differentiation yields which indicates that h 1 /h 2 is increasing in p by Lemma 1.
The following theorem gives a sufficient condition for the function F p (x) to be increasing on (0, π/2).
Finally, we consider the monotonicity of the function F p (x) on (0, π). In this case, we have to assume that 0 < p ≤ 1/π 2 .