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Math Help - Prove that it's strictly increasing

  1. #1
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    Prove that it's strictly increasing

    Prove that \left(1+\frac 1n\right)^n\left(1+\frac 1{4n}\right) is strictly increasing for n>1. Or in other words the recurrence equation is strictly increasing.
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    Moo
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    Hello,

    a_n=\left(1+\frac 1n\right)^n\left(1+\frac 1{4n}\right)

    Let's show that \frac{a_{n+1}}{a_n} > 1

    a_n=\frac{(n+1)^n}{n^n} \frac{4n+1}{4n}

    a_{n+1}=\left(1+\frac{1}{n+1} \right)^{n+1}\left(1+\frac 1{4(n+1)}\right)=\frac{(n+2)^{n+1}}{(n+1)^{n+1}} \frac{4n+5}{4(n+1)}

    \frac{a_{n+1}}{a_n}=\dfrac{(n+2)^{n+1} \overbrace{(4n+5)}^{\text{i \ take \ this \ one}}}{(n+1)^{n+1} 4(n+1)} \dfrac{\overbrace{n^n 4n}^{4n^{n+1}}}{(n+1)^n \underbrace{(4n+1)}_{\text{i \ take \ this \ one}}}

    \frac{a_{n+1}}{a_n}=\dfrac{4n+5}{4n+1} \dfrac{(n+2)^{n+1}n^{n+1}}{(n+1)^{2n+2}}=\dfrac{4n  +5}{4n+1} \dfrac{(n^2+2n)^{n+1}}{(n^2+2n+1)^{n+1}}


    And i have a problem oO
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    If

    Quote Originally Posted by james_bond View Post
    Prove that \left(1+\frac 1n\right)^n\left(1+\frac 1{4n}\right) is strictly increasing for n>1. Or in other words the recurrence equation is strictly increasing.
    you get too encumbered in the \frac{a_{n+1}}{a_n} method you could differntiate and see if it is monotonically increasing on (1,\infty_...haha
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  4. #4
    Moo
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    I derivated but it didn't give something interesting
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Desole

    Quote Originally Posted by Moo View Post
    I derivated but it didn't give something interesting
    But what do you mean it didnt give you something interesting?
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  6. #6
    Moo
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    You have to study the sign of the derivative. Well, good luck ^^ I've found it easier to study \frac{a_{n+1}}{a_n} but can't conclude (just too lazy maybe)
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Haha

    Quote Originally Posted by Moo View Post
    You have to study the sign of the derivative. Well, good luck ^^ I've found it easier to study \frac{a_{n+1}}{a_n} but can't conclude (just too lazy maybe)
    You are probably right it is much easier
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  8. #8
    Senior Member JaneBennet's Avatar
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    Let N=\left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{4n}  \right).

    Then \ln{N}=n\ln{\left(1+\frac{1}{n}\right)}+\ln{\left(  1+\frac{1}{4n}\right)} and N is strictly increasing \Leftrightarrow \ln{N} is strictly increasing.

    Let \mathrm{f}(x)=x\ln{\left(1+\frac{1}{x}\right)}+\ln  {\left(1+\frac{1}{4x}\right)}

    Then \mathrm{f}'(x)=\ldots=\ln{\left(1+\frac{1}{x}\righ  t)}+\frac{3}{x+1}-\frac{1}{x}

    This is positive for all x>1 because x>1 \Rightarrow both \ln{\left(1+\frac{1}{x}\right)}>0 and \frac{3}{x+1}-\frac{1}{x}>0.

    Hence \ln{N} is strictly increasing.
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