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2002
2002 11
¢
17(2) and (3). Proof
Set
1 1
e =1+1+ + ... + ,n∈ N+.
n 2! n!
For any given n ∈ N+, because
1 1 1
e + = e + + + ···+ , (i)
n p n (n +1)! (n +2)! (n + p)!
n +2 1
where p ∈ N+,p≥ 3, and < ,j =3, ···,p,wemayget
(n + j)! (n + j − 1)!
1 1 1 n +2 n +2
e + − e = + + ···+
n p n (n +1)! n +2 (n +1)! (n +3)! (n + p)!
1 1 1 1 1 1
< + + ···+ +
(n +1)! n +2 (n +1)! (n +2)! (n + p − 1)! (n + p)!
1 1
= + (e + − e ).
(n +1)! n +2 n p n
Thus, there holds the following inequality
1 1
e + − e < + (e + − e ). (ii)
n p n (n +1)! n +2 n p n
Furthermore, (ii) implies that
n +2
e + − e < . (iii)
n p n (n +1)!(n +1)
Notice that en+p → e (as p →∞). Letting p →∞in (iii). Then, there holds the inequality
n +2 1
e − e ≤< , (v)
n (n +1)!(n +1) n!n
which is the second inequality in (2). The first inequality is easy.
Set θn = n!n(e − en). From (2), it is easy to see that (3) is true.
3 (
Ideas of Proof:
360◦
(1). First, to give the formula for S : S =2n · 3 · sin ;
n n 2n · 3
(2). By using continuously the formula sin (2x)=2sinx cos x, to give the equality: sin 60◦=
60◦ 60◦ 60◦ 60◦
n−1 · ···
2 cos cos 2 cos −1 sin −1 ;
2 2 2n 2n √
360◦ 60◦ 60◦ 60◦ 3
(3). Notice that the facts sin =sin ,and1> cos ≥ co