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Bernoulli's inequality states

  1. Jan 13, 2005 #1


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    Where is the mistake in these few simple steps ?!

    First, I guess that

    [tex]\lim_{n\rightarrow \infty}\frac{n}{6^n}=0[/tex]

    I'll show it using the definition

    [tex]\left|\frac{n}{6^n}\right|<\epsilon \Leftrightarrow \frac{n}{6^n}<\epsilon[/tex]

    But, Bernoulli's inequality states that [itex](1+x)^n \geq 1+xn \ \forall x>-1[/itex] and [itex]\forall n \in \mathbb{N}[/itex]. So, with x = 5, I get that

    [tex]6^n \geq 1+5n \Rightarrow \frac{n}{6^n}\leq \frac{n}{1+5n}[/tex]


    [tex]\frac{n}{1+5n} < \epsilon \Leftrightarrow n<\epsilon + 5\epsilon n \Leftrightarrow n> \frac{\epsilon}{1-5\epsilon}[/tex]

    So basically, for all n greater than that, the inequality should be satisfied. But this is not so evidently.
  2. jcsd
  3. Jan 13, 2005 #2


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    Your steps were correct, but Bernoulli's inequality will not help you here.

    [tex]\frac{n}{6^n}\leq \frac{n}{1+5n}[/tex]

    Is correct, but you can't use it, since the right hand side obviously does not approach zero.
    It's basically the squeeze theorem. If you want to do it by comparison, you need something bigger than [itex]\frac{n}{6^n}[/itex] that will still converge to 0.
    Last edited: Jan 13, 2005
  4. Jan 13, 2005 #3
    You can also seek to prove the convergence of the series where the nth coefficient is given by n6^(-n). Ratio test does it.
  5. Jan 13, 2005 #4
    I think quasar987 was trying to prove the convergence on a sequence not on series.
    instead of writing (1+5)^n, it's better (5+1)^n or 6^n>=
    >= 5^n+n5^(n-1)+... or conveniently 6^n>n5^(n-1) because the term n5^(n-1) in closer to 6^n than 1+5n.
    finally=> (n/6^n)<(n/n5^(n-1)=(1/5^(n-1)< e =>
    n>=((lne^-1)/ln5) +1
  6. Jan 13, 2005 #5
    let a_n=n*6^(-n).

    lim a_n/a_n+1=1/6<1.

    Therefore the series converges.

    THEREFORE, the desired limit is 0. Done. It's called the test for divergence; look it up.
  7. Jan 14, 2005 #6


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    There is a similar test for sequences too

    [tex]\lim_{n \rightarrow \infty} \left|\frac{x_{n+1}}{x_n}\right|= L [/tex]

    and if L<1, lim = 0

    if L>1, lim = infinity.

    By the way, I still don't understand why it doesn't work... I'll comment on Galileo's post tonight when I have more time.
  8. Jan 16, 2005 #7


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    If I wanted to show that

    [tex]\lim_{n \rightarrow \infty}\frac{n}{1+5n}=0[/tex]

    This is not true, so I should not be able to find the N, right?

    But if the last line of my first post is correct, it is precisely saying I have found that N in the person of


    , no?

    So are you saying that whenever I use the argument of saying "this is smaller than that, therefor if I find the N for "that", it will be valid for "this" too", I have first to verify that "that" has the same limit as I'm trying to show "this" has? The limit evaluating process would never end!
  9. Jan 16, 2005 #8
    The last sequence of equivalences in your first post does not seem to be true when [itex]\epsilon < \frac{1}{5}[/itex].

    [tex]n < \epsilon + 5 \epsilon n \Leftrightarrow n - 5 \epsilon n < \epsilon \Leftrightarrow n(1 - 5 \epsilon) < \epsilon .[/tex]

    But if [itex]\epsilon < 1/5[/itex], then [itex]1 - 5 \epsilon > 0[/itex], so that

    n < \frac{\epsilon}{1 - 5 \epsilon}.
  10. Jan 16, 2005 #9
    Yes, use this. I think L=1/6 in this case. I made a typo in the above: should have been a_(n+1)/a_n.

    Now if you can't use this result directly, you can kind of use it indirectly...

    I think you'll be able to use that result to derive a formula resembling this:
    |a_n|<=(1/6)^n |a_0|.

    Now it may look a bit different. The exponent may be n+1, n+2, or n-1 or n-2. The idea is that you are squeezing a_n's absolute value to be smaller than something that is "clearly" going to 0.

    The other thing is that this may hold for n after a certain n', like for n>5 or something.

    Sorry I haven't worked out the details but I think that's how it will fall into place.
  11. Jan 18, 2005 #10


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    This N that I found, suposedly proving that the limit is 0, has been found without the use of any "squeeze", but only by algebraic manipulations on hypothesis expression.

    (the hypothesis expression is that [tex]\frac{n}{1+5n}<\epsilon[/tex])

    Yet, for epsilon = 1/50, for exemple, this N fails.

    How can I be sure after this, that

    [tex]\lim_{n \rightarrow \infty}\frac{1}{n}=0[/tex]

    , since this result is also proved, using the definition of limit, by simple algebraic manipulations on the hypothesis expression?
  12. Jan 19, 2005 #11
    See post #8, where I showed you where the problem is. The simple algebraic manipulations you did are wrong...
  13. Jan 19, 2005 #12


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    whenever in the process of proving a limit and you compare the terms with something else you are appealing to the squeeze theorem.

    If [itex]|a_n|<|b_n|[/itex] for all n>N for some N, and [itex]b_n \to 0 [/itex] as [itex]n \to \infty[/itex] then [itex]a_n \to 0[/itex] as [itex]n \to \infty[/itex].

    Now, what if [itex]|a_n|<|b_n|[/itex], but [itex]b_n[/itex] does not go to zero?
    Then we can't make any statement about the converge of [itex]a_n[/itex].
    That's what you have:

    [tex]\frac{n}{6^n}\leq \frac{n}{1+5n}[/tex]

    but [itex]\frac{n}{1+5n}[/itex] does not approach zero, so you can't use Bernouilli to estimate your original sequence.
  14. Jan 19, 2005 #13


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    Mmh, thanks for pointing that out Galileo. I hadn't noticed this is what we were doing even when we prove that a limit is something different then 0.

    Muzza, sorry about that, I thought your post #8 was only meant to show that my N fails for epsilon < 1/5. However, I don't understand what do you mean by

    What does it matter what the value of [itex]1 - 5 \epsilon > 0[/itex] is? Starting from the hypothesis expression that concerns any epsilon at all, you have arrived at an equivalent form, that is [itex]n(1 - 5 \epsilon) < \epsilon[/itex]. Isn't the natural and valid for all epsilon (except 1/5) next step

    [tex]n < \frac{\epsilon}{1 - 5 \epsilon}.[/tex]?!

    However, I do realise that the very fact that [itex]n < \frac{\epsilon}{1 - 5 \epsilon}.[/itex] explodes in my face for epsilon = 1/5 implies that the hypothesis expression is wrong for epsilon = 1/5, and therfor, wrong overall. And also, for epsilon < 1/5, [itex]n < \frac{\epsilon}{1 - 5 \epsilon}.[/itex] is true only for n < 1, therefor, for no n, since n is natural and so there exists no N. (perhaps this is what you meant?)

    There is one last thing that bothers me though... starting from [itex]n < \epsilon + 5 \epsilon n[/itex], I proceeded in a different way, and I don't see that's wrong with it..

    [tex]n < \epsilon + 5 \epsilon n \Leftrightarrow 0<\epsilon + 5 \epsilon n - n \Leftrightarrow 0< \epsilon + n(5\epsilon-1) \Leftrightarrow -n(5\epsilon-1)<\epsilon \Leftrightarrow n(5\epsilon-1)>-\epsilon[/tex]
    [tex] \Leftrightarrow n>\frac{-\epsilon}{5\epsilon-1} \Leftrightarrow n>\frac{\epsilon}{1-5\epsilon}[/tex]
  15. Jan 19, 2005 #14
    Aren't you familiar with how multiplication by a negative number affects an inequality? As in, if you multiply (or divide (same thing)) by a negative number, you must switch the inequality sign, if you multiply by a positive number you can leave it alone?

    If 1 - 5e > 0, the inverse of (1 - 5e) is also positive, so one can multiply both sides of n(1 - 5e) < e by 1/(1 - 5e) to get n < e/(1 - 5e).

    But if (1 - 5e) < 0, you must switch the signs around, so you get n > e/(1 - 5e).

    Thus, your original argument only holds when (1 - 5e) < 0 <=> e > 1/5...

    The problem with your other argument is practically the same as the first. You carelessly divide by something (i.e (5e - 1)) which might be negative!
    Last edited: Jan 19, 2005
  16. Jan 19, 2005 #15


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    AAAaaah! Phew, Muzza saves math, once again! :biggrin:
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