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Math Help - Bounding an sequence

  1. #1
    Member mohammadfawaz's Avatar
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    Bounding an sequence

    Hello all,

    I have the following sequence: a_n(x)=n(x^{\frac{1}{n}}-1) where x>1. The objective is to prove that it is bounded above and below. Also, we need to prove that it has no smallest element. I succeeded in proving that it is bounded below: (let x=1+h with h>0 and continue from there). For the other questions, I think it would be enough if we can prove that the sequence is decreasing. The upper bound would be simply a_1 = x-1. I tried to prove this fact but i reached: a_{n+1}-a_{n}=n(x^{\frac{1}{n+1}}-x^{\frac{1}{n}})+x^{\frac{1}{n+1}}-1. But I'm stuck there. Help please.

    Thanx
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mohammadfawaz View Post
    Hello all,

    I have the following sequence: a_n(x)=n(x^{\frac{1}{n}}-1) where x>1. The objective is to prove that it is bounded above and below. Also, we need to prove that it has no smallest element. I succeeded in proving that it is bounded below: (let x=1+h with h>0 and continue from there). For the other questions, I think it would be enough if we can prove that the sequence is decreasing. The upper bound would be simply a_1 = x-1. I tried to prove this fact but i reached: a_{n+1}-a_{n}=n(x^{\frac{1}{n+1}}-x^{\frac{1}{n}})+x^{\frac{1}{n+1}}-1. But I'm stuck there. Help please.

    Thanx
    Let z=\frac{1}{n} then we have \frac{x^z-1}{z} but x=1+\varepsilon and so \frac{(1+\varepsilon)^z-1}{z}\leqslant\frac{z\cdot\varepsilon+1-1}{z}=\varepsilon
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  3. #3
    Member mohammadfawaz's Avatar
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    But when using Bernoulli's inequality, you have to use > and not <. In this case, we get the lower bound which is \epsilon
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mohammadfawaz View Post
    But when using Bernoulli's inequality, you have to use > and not <. In this case, we get the lower bound which is \epsilon
    Not for the case when 0\leqslant z\leqslant 1.

    Theorem: Let 0\leqslant \delta\leqslant 1. Then, (1+x)^\delta\leqslant 1+\delta x.

    Proof: Define f(x)=1+\delta x-(1+\delta)^x. Clearly, f(0)=0 and f'(x)=\delta-\delta(1+x)^{\delta-1}\geqslant\delta-\delta=0. The conclusion follows.
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    Quote Originally Posted by mohammadfawaz View Post
    But when using Bernoulli's inequality, you have to use > and not <. In this case, we get the lower bound which is \epsilon

    Here is a proof :

    PlanetMath: proof of Bernoulli's inequality
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