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Math Help - Sequences question

  1. #1
    s3a
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    Sequences question

    Is the solution I posted wrong?

    For the last part, using L'Hopital's rule on (1 - xln(2))/2^x gives me -ln(2)/[ln(2) * 2^x] = -1/2^x = 0 (ignoring the limit operators - is operator the right word?).

    Is the solution wrong or am I not seeing something?

    Any input would be appreciated!
    Thanks in advance!
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Sequences question

    Quote Originally Posted by s3a View Post
    Is the solution I posted wrong?

    Using L'Hopital's rule on (1 - xln(2))/2^x gives me -ln(2)/[ln(2) * 2^x] = -1/2^x = 0 (ignoring the limit operators - is operator the right word?).

    Is the solution wrong or am I not seeing something?

    Any input would be appreciated!
    Thanks in advance!

    We are not megicians... WHAT is the question?!
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  3. #3
    s3a
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    Re: Sequences question

    The question is included with the solution.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Sequences question

    Quote Originally Posted by s3a View Post
    For the last part, using L'Hopital's rule on (1 - xln(2))/2^x gives me -ln(2)/[ln(2) * 2^x] = -1/2^x = 0 (ignoring the limit operators - is operator the right word?).
    You have to compute \lim_{x\to +\infty}\dfrac{x}{2^x} , an differentiate independently numerator and denominator.


    Out of curiosity, I don't understand why the problem uses the function f(x) , it is not necessary. For example for (a) s_{n+1}-s_n=\ldots=\dfrac{29}{(7n+10)(7n+3)}\geq 0 for all n which implies s_n is increasing and its limit is easily obtained by elemental computations. For (b) , \dfrac{s_{n+1}}{s_n}=\ldots=\dfrac{n+1}{2n}\leq 1 which implies s_n is decreasing etc.
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