数学系2002级三、四班数学分析补充材料(三) (单变元函数极限、单变元连续函数习题,2002年11月) 部分难题解析 补充题(一) 17(2)and(3).Pr any given n E N+, because En+p= en+ m+1)+(n+2 n+p)! where pEN+ P23, and (n+) <(n+i-1): 3=3,, P, we may get n+2 (n+1)n+21(n+1)(n+3 (n+p)! (n+1)!n+2(m+1)!(n+2)!"(n+p-1)!(n+p) (n+1)!n+ Thus, there holds the following inequ (n+1)!n+2 En+pen Furthermore, (ii) implies that n+2 (n+1)!(n+1) Notice that en+p→e(asp→∞). Letting p→oin(i).Then, there holds the inequality e- en ≤ (n+1)(n+1) which is the second inequality in(2). The first inequality is easy Set 0n=n!n(e-en). From(2), it is easy to see that(3)is true 习题1.3的第3题(见《简明数学分析》) (1). First, to give the formula for S:Sn=2n·3·sin360° (2)By using continuously the formula sin(2 z)=2 sin r cos ar, to give the equality: sin 60 2n-]sIn (3). Notice that the facts sin 10o 60° -and1>c0s可≥ Co It may be obtained: sin 2n.3 2 2n-11
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  17(2) and (3). Proof  Set en =1+1+ 1 2! + ... + 1 n! , n ∈ N+. For any given n ∈ N+, because en+p = en + 1 (n + 1)! + 1 (n + 2)! + ··· + 1 (n + p)!, (i) where p ∈ N+, p ≥ 3, and n + 2 (n + j)! < 1 (n + j − 1)!, j = 3, ··· , p, we may get en+p − en = 1 (n + 1)! + 1 n + 2
1 (n + 1)! + n + 2 (n + 3)! ··· + n + 2 (n + p)! < 1 (n + 1)! + 1 n + 2
1 (n + 1)! + 1 (n + 2)! ··· + 1 (n + p − 1)! + 1 (n + p)! = 1 (n + 1)! + 1 n + 2 (en+p − en). Thus, there holds the following inequality en+p − en < 1 (n + 1)! + 1 n + 2 (en+p − en). (ii) Furthermore, (ii) implies that en+p − en < n + 2 (n + 1)!(n + 1). (iii) Notice that en+p → e (as p → ∞). Letting p → ∞ in (iii). Then, there holds the inequality e − en ≤ n + 2 (n + 1)!(n + 1) < 1 n!n, (v) 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.  1.3  3  ( Ideas of Proof: (1). First, to give the formula for Sn: Sn = 2n · 3 · sin 360◦ 2n · 3 ; (2). By using continuously the formula sin (2x) = 2 sin x cos x, to give the equality: sin 60◦ = 2n−1 cos 60◦ 2 · cos 60◦ 22 ··· cos 60◦ 2n−1 sin 60◦ 2n−1 ; (3). Notice that the facts sin 360◦ 2n · 3 = sin 60◦ 2n−1 , and 1 > cos 60◦ 2j ≥ cos 60◦ 2 = √3 2 , j = 1, 2, ..., n − 1. It may be obtained: sin 360◦ 2n · 3 < √3 2 · 1 2n−1 · 2 √3 n−1 = √3 2 · 1 √3 n−1 .