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Math Help - Proof via MVT

  1. #1
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    Proof via MVT

    For x>0, (1/x^2) - 1/(x+1)^2 >= 2/(x+1)^3 and therefore, for any positive integer n:
    (1/n^2) - 1/(n+1)^2 >= 2/(n+1)^3

    Prove this fact using the Mean Value Theorem.

    Well, I suppose we should start by making f(x)=2/(x+1)^3
    and so the MVT says that there is some c so that
    f'(c)=f(t)-f(0)/(t-0) and then f'(c)*t+f(0)=f(t), but I don't know, I'm confused. This is how most of the other proofs in the text start off....
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    For x>0, (1/x^2) - 1/(x+1)^2 >= 2/(x+1)^3 and therefore, for any positive integer n:
    (1/n^2) - 1/(n+1)^2 >= 2/(n+1)^3
    Prove this fact using the Mean Value Theorem.
    Well, I suppose we should start by making f(x)=2/(x+1)^3
    No start with f(x)=(x)^{-2}.
    Consider the interval [n,n+1]
    Last edited by Plato; November 26th 2009 at 09:29 AM.
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  3. #3
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    Ok. So then we have that there is c in this interval so

    f'(c)=f(t+1)-f(t)/(t+1)-t=f(t+1)-f(t)=(1/t^2+1)-(1/t^2) =1.

    Now clearly 2/(x+1)^3<1 for x>=1

    Is that it?
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  4. #4
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    This is hopeless. What do you understand about this problem?
    If f(x)=x^{-2} then f'(x)=-2x^{-3}.
    By the MVT \left( {\exists c \in (n,n + 1} \right)\left[ {f'(c) = \frac{{ - 2}}{{c^3 }} = \frac{1}{{(n + 1)^2}} - \frac{1}{n^@}} \right].

    Simple algebra gives us \frac{2}{{c^3 }} = \frac{1}{n^2} - \frac{1}{{(n + 1)^2}}
    and also n < c < n + 1\, \Rightarrow \,n^3  < c^3  < \left( {n + 1} \right)^3 \, \Rightarrow \,\frac{1}{{\left( {n + 1} \right)^3 }} < \frac{1}<br />
{{c^3 }} < \frac{1}{{n^3 }}.
    At this point allow me to be perfectly frank.
    You in the past posted questions without any evidence of effort on your part.
    So at this point I must tell you that if you cannot finish this question given this extensive advice, then you must simply say to yourself “I am drowning in my own misunderstanding”.
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