✔ 最佳答案
Can try MI:
Let P(n) be the statement:
Σ(k = 1 → n) 1/√k < √(n + 1) + √n - 1
When n = 1,
LHS = 1
RHS = √2
So P(1) is true.
Assume that P(r) is true, where r is a positive integer, i.e.
Σ(k = 1 → r) 1/√k < √(r + 1) + √r - 1
Adding 1/√(r + 1) to both sides:
Σ(k = 1 → r + 1) 1/√k < √(r + 1) + √r + 1/√(r + 1) - 1
Comparing √(r + 1) + √r + 1/√(r + 1) and √(r + 2) + √(r + 1):
[√(r + 1) + √r + 1/√(r + 1)] - [√(r + 2) + √(r + 1)]
= [√r - √(r + 2)] + 1/√(r + 1)
= [√r - √(r + 2)][√r + √(r + 2)]/[√r + √(r + 2)] + 1/√(r + 1)
= -2/[√r + √(r + 2)] + 1/√(r + 1)
Now, from:
2√[r(r + 2)] < r + r + 2 = 2(r + 1) (AM > GM)
2r + 2 + 2√[r(r + 2)] < 4(r + 1)
[√r + √(r + 2)]2 < [2√(r + 1)]2
√r + √(r + 2) < 2√(r + 1)
1/√(r + 1) < 2/[√r + √(r + 2)]
-2/[√r + √(r + 2)] + 1/√(r + 1) < 0
[√(r + 1) + √r + 1/√(r + 1)] < [√(r + 2) + √(r + 1)]
Σ(k = 1 → r + 1) 1/√k < √(r + 2) + √(r + 1) - 1
Hence P(r + 1) is also true.
By the principle of MI, P(n) is true for all positivei integers n.
2010-01-28 23:23:18 補充：
Yes, this is called the method of synthesis
