证明:1/f(1)+1/f(2)+1/f(3)+...+1/f(n)

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证明:1/f(1)+1/f(2)+1/f(3)+...+1/f(n)

证明:1/f(1)+1/f(2)+1/f(3)+...+1/f(n)
证明:1/f(1)+1/f(2)+1/f(3)+...+1/f(n)<4/3
f(n)=n^3-(n-1)^3

证明:1/f(1)+1/f(2)+1/f(3)+...+1/f(n)
1/ ((n+1)^3-n^3)= 1/(3n^2+3n+1) < 1/3 * 1/(n^2+n)= 1/3 (1/n - 1/(n+1))
===>
1/f(1)+1/f(2)+1/f(3)+...+1/f(n)
< 1/f(1) + 1/3 (1/1 - 1/2 + .+ 1/(n-1) - 1/n)
= 1 + 1/3 - 1/(3n) < 4/3

f(n)=n^3-(n-1)^3=3n^2-3n+1,当n>=1时,函数单调递增,因此恒为正。于是有:
1/f(1)+1/f(2)+1/f(3)+...+1/f(n)等于对1/(3k^2-3k+1)从k=1到k=n求和。为了表达方便,此处用{1/(3k^2-3k+1)@(m,n)}表示对1/(3k^2-3k+1)从k=m到k=n求和,则:
{1/(3k^2-3k+1)@(1,n)}...

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f(n)=n^3-(n-1)^3=3n^2-3n+1,当n>=1时,函数单调递增,因此恒为正。于是有:
1/f(1)+1/f(2)+1/f(3)+...+1/f(n)等于对1/(3k^2-3k+1)从k=1到k=n求和。为了表达方便,此处用{1/(3k^2-3k+1)@(m,n)}表示对1/(3k^2-3k+1)从k=m到k=n求和,则:
{1/(3k^2-3k+1)@(1,n)} = 1+{1/(3k^2-3k+1)@(2,n)}< 1+{1/(3k^2-3k)@(2,n)}=1+{(1/3)[1/(n-1)-1/n]@(2,n)}=1+(1/3){[1/(n-1)-1/n]@(2,n)}=1+(1/3)(1-1/n)<1+(1/3)=4/3
OVER!

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