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Math Help - Proof

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Proof

    I have seen the proof of this, but I would like to see if anyone can do it easier..


    prove that

    \forall{n}>0,\text{  }\ln\bigg(1+\frac{1}{n}\bigg)>\frac{2}{2n+1}
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  2. #2
    MHF Contributor arbolis's Avatar
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    I didn't try, but maybe by induction?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I didn't try, but maybe by induction?
    The proof I have seen involves integrals and infinite series, induction might work, but I am not entirely sure that would be easier
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    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    I have seen the proof of this, but I would like to see if anyone can do it easier..


    prove that

    \forall{n}>0,\text{ }\ln\bigg(1+\frac{1}{n}\bigg)>\frac{2}{2n+1}
    Put:

     <br />
f(x)=\ln\left(1+\frac{1}{x}\right)-\frac{2}{2x+1}<br />

    It is fairly easy to show that f(x) is strictly decreasing for x > 0, and that the limit as x \to \infty is 0. Which implies that f(x) has no roots for x > 0 and as it is greater than zero for arbitarily small x that proves your result.

    RonL
    Last edited by CaptainBlack; June 3rd 2008 at 08:15 PM.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Put:

     <br />
f(x)=\ln\left(1+\frac{1}{x}\right)-\frac{2}{2x+1}<br />

    It is fairly easy to show that f(x) is strictly decreasing for x > 0, and that the limit as x \to \infty is 0. Which implies that f(x) has no roots for x > 0 and as it is greater than zero as for arbitarily small x that proves your result.

    RonL
    Yeah, that is a well known method of showing that an inequality is true, but can you use a similar method to find that other function.

    For example this method shows the inequality, but you cannot not use it to go from

    \ln\bigg(1+\frac{1}{x}\bigg)>f(x)

    then f(x)=\frac{2}{2x+1}

    I guess I wasnt clear enough in my post, sorry
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    Yeah, that is a well known method of showing that an inequality is true, but can you use a similar method to find that other function.

    For example this method shows the inequality, but you cannot not use it to go from

    \ln\bigg(1+\frac{1}{x}\bigg)>f(x)

    then f(x)=\frac{2}{2x+1}

    I guess I wasnt clear enough in my post, sorry
    Now this is not clear, what do you wish to show?.

    I have shown that:

    <br />
\ln\left(1+\frac{1}{n}\right)>\frac{2}{2n+1}<br />

    for all n>0, which was the original question.

    RonL
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Now this is not clear, what do you wish to show?.

    I have shown that:

    <br />
\ln\left(1+\frac{1}{n}\right)>\frac{2}{2n+1}<br />

    for all n>0, which was the original question.

    RonL
    I am sorry, I will state explicitly what I want. Lets say that I was just given \ln\bigg(1+\frac{1}{n}\bigg) and I needed to find some function that is always smaller than it but will also diverge by the integral test. In other words your method shows how to show that the inequality is true. But in only works given the fact taht we know the inequality, say we didnt know the function that was smaller than \ln\bigg(1+\frac{1}{n}\bigg) and diverged by the integral test. How would one go about taking \ln\bigg(1+\frac{1}{n}\bigg) and manipulating it to find such a function.

    If that was convoluted just ignore it, there is no need to prove this, its just out of curiosity
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    prove that

    \forall{n}>0,\text{ }\ln\bigg(1+\frac{1}{n}\bigg)>\frac{2}{2n+1}
    for this particular problem CaptainBlack's solution is probably the easiest, but the problem is a very special case

    of the following beautiful result: let f: [a,b] \rightarrow \mathbb{R} be such that f'''(x) exists on [a,b]. we know that by the mean

    value theorem f(b)-f(a)=(b-a)f'(c). now it can be proved that:

    1) if f''(x)f'''(x) < 0, \ \forall x \in [a,b], then a < c < \frac{a+b}{2}.

    2) if f''(x)f'''(x) > 0, \ \forall x \in [a, b], then \frac{a+b}{2} < c < b.

    the proposer, i guess, was professor Alex Lupas, a Romanian mathematician who sadly passed away few months

    ago. my proof of this result is easy to follow and i can post it here, but some of you might want to think about it

    a little bit first?
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    for this particular problem CaptainBlack's solution is probably the easiest, but the problem is a very special case

    of the following beautiful result: let f: [a,b] \rightarrow \mathbb{R} be such that f'''(x) exists on [a,b]. we know that by the mean

    value theorem f(b)-f(a)=(b-a)f'(c). now it can be proved that:

    1) if f''(x)f'''(x) < 0, \ \forall x \in [a,b], then a < c < \frac{a+b}{2}.

    2) if f''(x)f'''(x) > 0, \ \forall x \in [a, b], then \frac{a+b}{2} < c < b.

    the proposer, i guess, was professor Alex Lupas, a Romanian mathematician who sadly passed away few months

    ago. my proof of this result is easy to follow and i can post it here, but some of you might want to think about it

    a little bit first?
    I dont want to look stupid(I probably will end up doing so) but, this does not make sense to me unless it should be

    f\bigg(\frac{a+b}{2}\bigg)<f(c)<f(b) due to the fact of the concavity etc.
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  10. #10
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    Quote Originally Posted by Mathstud28 View Post
    I dont want to look stupid(I probably will end up doing so) but, this does not make sense to me unless it should be

    f\bigg(\frac{a+b}{2}\bigg)<f(c)<f(b) due to the fact of the concavity etc.
    i guess you're reffering to part 2) of the result? the result is as i mentioned, i.e. if f''f''' > 0 over the interval,

    then \frac{a+b}{2} < c < b. the beauty of the result is that it gives a better location of c. the mean value theorem can

    only tell us that a < c < b. but under those conditions that i mentioned, we can tell if c is in the first half of the

    interval or in the second half. probably it'd help if you see it in a couple of examples.
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