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Thread: Double Induction for (1+t)^t > 1+nt

  1. #1
    Jan 2011

    Double Induction for (1+t)^t > 1+nt

    Hi, I have the following problem which I think I have solved correctly (it would be great if you could check it), but I don't know how to answer to "simple" questions about my proof... any help would be very much appreciated, thanks!

    The problem says:
    "Let t>-1 be given with t≠0. Show by induction that for n≥2 we have
    $\displaystyle (1+t)^t>1+nt$."

    And here are the two questions that I don't know:

    "Where is the condition t>-1 used in your proof? Is the result true if instead we had -2<t<-1?"

    My solution to the first bit is:

    Let $\displaystyle F(n)=(1+t)^n>1+nt$.

    Proof by induction on n.

    Base n=0. Then $\displaystyle (1+t)^n=(1+t)^0=1=1+0t=1+nt$

    Induction. Suppose that there is a number k (k≥2) such thaqt $\displaystyle (1+t)^k>1+kt$ for any t>-1 (t≠0). We need to show that (1+t)^{k+1}>1+(k+1)t for any t>-1, t≠0.

    $\displaystyle (1+t)^{k+1}=(1+t)^k*(1+t)$. By the induction hypothesis, $\displaystyle (1+t)^k>1+kt$, so we have:
    $\displaystyle (1+t)^k*(1+t) \geq (1+kt)*(1+t)=1+kt+t+kt^2=1+(1+k)t+kt^2$.

    Since $\displaystyle t^2>0$ and $\displaystyle k \geq 2, 1+(1+k)t+kt^2 \geq 1+(1+k)t$.

    So $\displaystyle (1+t)^{k+1} \geq 1+(k+1)t$

    And so the result holds for n=0 and hence for all n≥2 and t>-1,t≠0.
    Last edited by jmgilbert; Mar 10th 2011 at 12:38 PM.
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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    Everything looks good, except for one thing: you skipped the n = 1 case! It's a small technicality, but it does actually render your proof invalid. You definitely have the right idea with your proof. I would clean up your notation a bit. Note: it's \geq or \ge, not /geq. That'll make your post a lot more intelligible.
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  3. #3
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Two comments. First, as Ackbeet says, you must start the induction at n=2, not n=0, if you want to get the strict inequality (for $\displaystyle t\ne0$).

    The second comment is that in the inductive step you multiplied both sides of an inequality by $\displaystyle (1+t)$. Now you ought to know that an inequality is preserved if you multiply both sides by a positive number, but not if you multiply both sides by a negative number. That should help you to answer the "simple" questions about the proof.
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