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Decreasing sequence (proof)

  1. Oct 23, 2006 #1
    i have to prove that the sequence {ak} is decreasing, where
    [tex]\{a_k\} = \{\left({1+\frac{1}{k}}\right)^{-k}\}[/tex]

    this is what i did:

    [tex]a_k = {\left(\frac{k}{k+1}}\right)^k[/tex]

    [tex]a_{k+1}-a_{k}[/tex]

    [tex]= {\left(\frac{k+1}{k+2}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)}^{k}[/tex]

    [tex]= {\left(\frac{1+\frac{1}{k}}{1+\frac{2}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}[/tex]

    [tex]< {\left(\frac{1+\frac{1}{k}}{1+\frac{1}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}[/tex]

    [tex]= 1 - \left({\frac{k}{k+1}}\right)^{k}[/tex]

    [tex]< 1-1[/tex] since [tex]\left({\frac{k}{k+1}}\right)^{k} < 1[/tex]

    therefore, [tex]a_{k+1}-a_{k} < 0[/tex]

    therefore, the sequence is decreasing.

    am i right?
     
    Last edited: Oct 23, 2006
  2. jcsd
  3. Oct 23, 2006 #2

    arildno

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    Hmm..no
    The RATIO between the terms should be less than 1, not the DIFFERENCE!
    The difference should be less than 0
     
  4. Oct 23, 2006 #3

    Office_Shredder

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    Suppose we take the sequence 1/2, 1/2, 1/2.

    ak+1 - ak = 0 < 1.

    Therefore, the sequence is decreasing :P
     
  5. Oct 23, 2006 #4
    ok, i made a stupid mistake. i have edited my proof. is it correct now?
     
    Last edited: Oct 23, 2006
  6. Oct 23, 2006 #5

    arildno

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    No, it is not.
    The reason is that since the last fraction is LESS than 1, the DIFFERENCE between 1 and the fraction is GREATER than the difference between 1 and 1.
    Your first bound is too crude to derive the result.
     
    Last edited: Oct 23, 2006
  7. Oct 23, 2006 #6
    I suggest the following:

    Put the expression under the form of e^f(k). The derivative of that expression would be e^f(k) * f'(k). Knowing that e^f(k) is positive for k sufficiently large, find the sign that f'(k) assumes for an also sufficiently k. If the sign is negative, the expression e^f(k) * f'(k) is negative for this value of k and beyond, and you have proven that e^f(k) is decreasing.

    Edit: you will find f(k) = ln (1 + 1/k) * - k. If y = ln (1 + 1/k), y' = -1/k(k+1) and f'(k) = -ky' - y = 1/(k+1) - ln (1 + 1/k).

    Now (unfortunatly), this is not relevant. We are confronted to another inequality, ln (1 + 1/k) > 1/(k+1).

    Sorry that I've wasted so much time... maybe the second inequality would help, but I doubt it. To compensate I'll give you another proof , we have:

    (1+1/k)^k is increasing, for

    [tex] (1+1/n)^{n} = 1 + n\frac{1}{n} + \frac{1}{2!}(1-\frac{1}{n})+ ... + \frac{1}{n!}(1 - \frac{1}{n})(1 - \frac{2}{n})...(1 - \frac{n-1}{n}) [/tex].

    This is simply by using the binominial theorem. Now if,

    [tex]S_n = 1 + n\frac{1}{n} + \frac{1}{2!}(1-\frac{1}{n})+ ... + \frac{1}{n!}(1 - \frac{1}{n})(1 - \frac{2}{n})...(1 - \frac{n-1}{n}) [/tex],

    then S_m > S_n. Since (1+1/m)^m = S_m + ...., then (1+1/m)^m > (1+1/n)^n

    and

    (1+1/m)^-m < (1+1/n)^-n

    Thus the expression is decreasing. It should be noted that this is only a proof for integrer values of n.
     
    Last edited: Oct 23, 2006
  8. Oct 23, 2006 #7
    is it correct?
    if [tex]y = \ln\left(1+\frac{1}{x}\right)[/tex] then i get [tex]y' = \frac{-1}{x(x+1)}[/tex]
     
    Last edited: Oct 23, 2006
  9. Oct 23, 2006 #8
    is there any way to prove that the sequence is decreasing the way i started?
     
    Last edited: Oct 23, 2006
  10. Oct 23, 2006 #9
    Wait, wait, wait, my proof is erroneous. [tex]y = \ln\left(1+\frac{1}{x}\right)[/tex] then [tex]y' = \frac{-1}{x(x+1)}[/tex]

    You are correct. I'll edit my proof right now.
     
    Last edited: Oct 23, 2006
  11. Oct 26, 2006 #10
    is there any way to prove the sequesnce is decreasing by one of the following:
    [tex]a_{k+1}-a_k < 0[/tex]

    [tex]\frac{a_{k+1}}{a_k} < 1[/tex]

    [tex]f'(k) < 0[/tex]
     
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