pure inequality 1條10分 o.o~~~~~

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

sign=(-1)^k ?