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[ Last edited by ÎÒÒª·É on 2012-10-2 at 22:16 ]
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Proof
¡Æ_(k=1)^¡Þ▒¡¼(-1)^(k-1) (k^(2-a)-(k-1)^(2-a) )=[1-0]-[ 2^(2-a)-1]+[3^(2-a)-2^(2-a)]-[4^(2-.a)-3^(2-a)]+⋯
=[2.1-(0+2^(2-a))]+[2.3^(2-a)-(2^(2-a)+4^(2-a))]+[2.5^(2-.a)-((4^(2-a)+6^(2-a))]+[2.7^(2-.a)-(6^(2-a)+8^(2-a)]+⋯
It is obvious the first term [2.1-(0+2^(2-a))]  is positive.  We¡¯ll prove every term in the square brackets is positive.
An arbitrary term in a square bracket can be expressed as
[2k^(2-a)-((k-1)^(2-a)+(k+1)^(2-a))], here k is an odd number.
Let¡¯s consider a function y=x^(2-a), x>0, 1 Since y¡¯=(2-a)x1-a > 0 and y¡¯¡¯=(2-a)(1-a)x-a < 0, y is a monotonically increasing concave function.
If you sketch a graph of y=x^(2-a) (I have the graph available, but I do not know how to post it), it is obvious that
[2k^(2-a)>((k-1)^(2-a)+(k+1)^(2-a))]
Or
[2k^(2-a)-((k-1)^(2-a)+(k+1)^(2-a))]>0
As a result, the original sum is positive.
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8Â¥2012-10-06 06:04:49
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3Â¥2012-10-03 17:35:08
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