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# prove it

x1,x2,x3,......,xn為互不相同的正整數,n>1

求證:1/x1^2+1/x2^2+1/x3^2+......+1/xn^2<2

Update:

x1,x2,x3,......,xn不一定是由1到n的正整數

例如2,4,7,10,......,150

### 1 Answer

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- Anonymous1 decade agoFavorite Answer
由於 x1,x2,x3,......,xn為互不相同的正整數, 換言之, 它們最小的可能值為 1, 2, 3, ... , n (不分次序)

所以:

1/[(x1)²] + 1/[(x2)²] + 1/[(x3)²] + ... 1/[(xn)²] 的最大可能值為:

1/1² + 1/2² + 1/3² + ... + 1/n²

= 1 + 1/2² + 1/3² + ... + 1/n²

現在:

1/2² + 1/3² + ... + 1/n²

= Σ (i = 2 → n) 1/i²

< Σ (i = 2 → n) 1/[i(i - 1)]

= Σ (i = 2 → n) [1/(i - 1) - 1/i]

= [Σ (i = 2 → n) 1/(i - 1)] - [Σ (i = 2 → n) 1/i]

= [Σ (i = 1 → n - 1) 1/i] - [Σ (i = 2 → n) 1/i]

= 1 - 1/n

< 1

所以,

1 + 1/2² + 1/3² + ... + 1/n² < 1 + 1

1 + 1/2² + 1/3² + ... + 1/n² < 2

即是 1/[(x1)²] + 1/[(x2)²] + 1/[(x3)²] + ... 1/[(xn)²] 的最大可能值仍然小於 2.

結論:

1/[(x1)²] + 1/[(x2)²] + 1/[(x3)²] + ... 1/[(xn)²] < 2

Source(s): My Maths knowledge

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