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Thread: Proof by Induction : Inequalities

  1. #1
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    Proof by Induction : Inequalities

    Proof by induction that
    $\displaystyle
    1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{n}}< 2\sqrt{n}

    $

    K+1 Step

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

    $

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

    Then im stuck here....

    Any help would be appreciated.
    Thanks in advance
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  2. #2
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    You left out a term in the last line. It should read

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

    What you need to show now is that:

    $\displaystyle \frac{2\sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}}\leq 2\sqrt{k+1}$,

    or equivalently, that

    $\displaystyle 2\sqrt{k} \sqrt{k+1}\leq 2k+1$.

    Squaring both sides (both sides are positive, as $\displaystyle k\in\mathbb N$) we get

    $\displaystyle 4k(k+1)\leq (2k+1)^2$.

    I'll let you conclude
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  3. #3
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    Thanks for help
    I think i get it

    so
    $\displaystyle 4k(k+1)\leq (2k+1)^2
    $

    Then

    $\displaystyle 4k^2+4k \leq 4k^2+4k+1 $

    That means

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

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

    So $\displaystyle K+1 \in\mathbb N $
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  4. #4
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    "so $\displaystyle K+1\in N$"? There was no "K" in what you did before. If you mean "k", that is not what you were trying to prove and there is no need to say it- since "k" was in N, certainly k+1 is.

    Your last line should be "so the statement is true for all n".
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  5. #5
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    Quote Originally Posted by firebio View Post
    Proof by induction that
    $\displaystyle
    1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{n}}< 2\sqrt{n}

    $

    K+1 Step

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

    $

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

    Then im stuck here....

    Any help would be appreciated.
    Thanks in advance
    There is no need to bring in $\displaystyle \le$

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

    Therefore, the following should be true

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

    We try to prove this using the first statement, since that will link every pair of adjacent natural numbers in a chain.

    $\displaystyle 2\sqrt{k}+\frac{1}{\sqrt{k+1}}=\frac{2\sqrt{k}\sqr t{k+1}}{\sqrt{k+1}}+\frac{1}{\sqrt{k+1}}=\frac{2\s qrt{k^2+k}+1}{\sqrt{k+1}}$

    Is $\displaystyle \frac{2\sqrt{k^2+k}+1}{\sqrt{k+1}}<2\sqrt{k+1}$ ?

    $\displaystyle 2\sqrt{k^2+k}+1<2(k+1)$ ?

    $\displaystyle 2\sqrt{k^2+k}<2k+1$ ?

    $\displaystyle \sqrt{4k^2+4k}<\sqrt{4k^2+4k+1}$ ?

    Yes.

    The adjacent-term chain is established.
    Finally prove the inequality is true for an initial value of n.
    Then it's true for that n and all following n.
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