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Thread: Induction Proof (Inequality) SIMPLE!!

  1. #1
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    Induction Proof (Inequality) SIMPLE!!

    Show 1/(n+1)+1/(n+2)+1/(n+3)+1/2n>3/5, for n>2. Use mathematical induction.

    THANKS
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  2. #2
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    Quote Originally Posted by GamesonPlanes View Post
    Show 1/(n+1)+1/(n+2)+1/(n+3)+1/2n>3/5, for n>2. Use mathematical induction.

    THANKS
    Is anything missing?

    If you put n=10 or more or so, the inequality will not hold
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  3. #3
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    I think the statement is correct. To prove it by induction, write the nth and (n+1)st sums: $\displaystyle s_n=1/(n+1)+\dots+1/(2n)$; $\displaystyle s_{n+1}=1/(n+2)+\dots+1/(2n+2)$. Then express $\displaystyle s_{n+1}$ as $\displaystyle s_n+f(n)$ where $\displaystyle f(n)$ is some expression of n, i.e., $\displaystyle f(n)=s_{n+1}-s_n$. If you show that $\displaystyle f(n)>0$, you are done.

    Note that $\displaystyle f(n)$ has three terms, and two of them are at least as big as half of the third.
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  4. #4
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    Quote Originally Posted by GamesonPlanes View Post
    Show 1/(n+1)+1/(n+2)+1/(n+3)+1/2n>3/5, for n>2. Use mathematical induction.

    THANKS
    Do you want to sum all terms from $\displaystyle \frac{1}{n+1}$ up to $\displaystyle \frac{1}{2n}$

    not just the 4 terms quoted?

    F(3)

    $\displaystyle \frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{9}{20}+\ frac{1}{6}=\frac{37}{60}>\frac{36}{60}$

    True for n=3

    F(k)

    $\displaystyle \frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+.....+\f rac{1}{2k}>\frac{3}{5}$ ?

    F(k+1)

    $\displaystyle \frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac {1}{2k+1}+\frac{1}{2k+2}$

    $\displaystyle =F(k)+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}$

    $\displaystyle =F(k)+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{2}{2k+2}$

    $\displaystyle =F(k)+\frac{1}{2k+1}-\frac{1}{2k+2}$

    A greater value in the denominator makes a smaller fraction, hence

    $\displaystyle \frac{1}{2k+1}-\frac{1}{2k+2}>0$

    Therefore if F(k) is true, F(k+1) certainly is
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