Observe that
Define function as follows:
Then we can rewrite (1.3) as
or
where is an odd integer and .
The following lemma can be obtained by simple calculations.
Lemma 2.1.
Let be defined by (2.2). Then
Lemma 2.2.
Assume that . If , then
where
.
Proof.
Since is symmetric in , we can assume, without loss of generality, that . Then there are possible cases:

(1)

(2)

(3)

(4)
(m+1)
And, for the above cases –(m+1), by the monotonicity of , in turn, we may get

(1)
;

(2)

(3)

(4)
(m+1).
From the above inequalities, it follows that (2.6) holds. The proof is complete.
Lemma 2.3.
Assume that . Then
.
Proof.
For , it is easy to see that
which yields
and so
It follows that (2.8) holds. Similarly, for , it is easy to see that
which yields
It follows that (2.9) holds. The proof is complete.
Lemma 2.4.
Let
where
Assume that Then
Proof.
By induction, we easily show that
It follows from Lemma 2.3 that
Hence, by (2.15) and (2.18), we have
Equation (2.20) implies that the limits and exist, and
It follows from (2.16) that
. Let in (2.15), we have
It follows that there exist such that
From (2.24), we have
Since
it follows from (2.25) and (2.18) that . The proof is complete.