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Math Help - Show 1/x+1 less than or equal to ln(x+1)/x

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    Show 1/x+1 less than or equal to ln(x+1)/x

    1/x+1 less than or equal to ln(x+1)/x
    using definition of ln(x) as intergral of 1 to x of 1/t dt

    show that for every real number we have x such that
    Last edited by mr fantastic; November 27th 2011 at 10:13 AM. Reason: Re-titled.
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    Re: analysis question please help?

    Quote Originally Posted by sbacon1991 View Post
    1/x+1 less than or equal to ln(x+1)/x
    using definition of ln(x) as intergral of 1 to x of 1/t dt
    show that for every real number we have x such that
    By definition: if x>0 then \ln (x) = \int_1^x {\frac{{dt}}{t}} .

    So \left( {\ln (x)} \right)^\prime   = \frac{1}{x} which means we can use the mean value theorem.

    If 0<a<b then \left( {\exists c \in (a,b)} \right)\left[ {\frac{1}{c} = \frac{{\ln (b) - \ln (a)}}{{b - a}}} \right]

    But a<c<b implies \frac{1}{b}<\frac{1}{c}<\frac{1}{a}

    That gives \frac{1}{b}<\frac{{\ln (b) - \ln (a)}}{{b - a}}}<\frac{1}{a}.

    Let a=1~\&~b=x+1 and we have it.
    Last edited by Plato; November 27th 2011 at 10:47 AM.
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