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Math Help - lim as n approach infinity

  1. #1
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    lim as n approach infinity

    Hello.

    I got stuck on this:

    Show using the binomial theorem, that if a>1 then \lim_{n\rightarrow\infty}a^n=+\infty

    Hereīs my brave attempt:

    (1+(a-1))^n=\sum_{k=1}^{n}\binom{n}{k}1^{n-k}(a-1)^k


    =\binom{n}{1}(a-1)+\binom{n}{2}(a-1)^2+...+\binom{n}{n}(a-1)^n

    So at first I thougt that each term is bigger than one (this is the part that I got wrong). Then, since there are n terms, the sum must tend to infinity because n tends to infinity and since each term bigger than one.

    The b part:

    Show that
    \lim_{n\rightarrow\infty}n^{-1}a^n=+\infty

    I want to try this myself after getting the a) part straight but I wonder is there a clue in the b part for solving a?

    Any hints, besides replacing my toilet paper with my home assignment?
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  2. #2
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    Quote Originally Posted by wilbursmith View Post
    Hello.

    I got stuck on this:

    Show using the binomial theorem, that if a>1 then \lim_{n\rightarrow\infty}a^n=+\infty

    Hereīs my brave attempt:

    (1+(a-1))^n=\sum_{k=1}^{n}\binom{n}{k}1^{n-k}(a-1)^k Not quite right – the sum should start with k=0, not k=1.

    =\binom{n}{1}(a-1)+\binom{n}{2}(a-1)^2+...+\binom{n}{n}(a-1)^n

    So at first I thougt that each term is bigger than one (this is the part that I got wrong). Then, since there are n terms, the sum must tend to infinity because n tends to infinity and since each term bigger than one.
    You have made a good start. So you have the equation a^n=1+\binom{n}{1}(a-1)+\binom{n}{2}(a-1)^2+...+\binom{n}{n}(a-1)^n. You don't need each term on the right to be greater than 1. You just need to say that each term is positive, so that a^n is greater than any single term on the right. In particular, a^n > \binom{n}{1}(a-1), and that tends to infinity as n does.

    Quote Originally Posted by wilbursmith View Post
    The b part:

    Show that
    \lim_{n\rightarrow\infty}n^{-1}a^n=+\infty

    I want to try this myself after getting the a) part straight but I wonder is there a clue in the b part for solving a?
    Yes there is. This time, try using one of the other terms on the right-hand side of that binomial expansion.
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  3. #3
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    Ok I think the a part is rather clear now. The only way that I see a^n is bigger than each term though is that a sum is always bigger than any of itīs terms assuming all positive (not much proof there). Is there a more convincing way of seeing that a^n actually is bigger than any of the terms?

    To the b part. I thought that \lim\frac{a^n}{n}\rightarrow\infty since numerator grows faster than denominator, but I quess my prof will repeat my name three times with decreasing pitch while shaking his head as he reads my conclusion.

    I wanted to do a binomial expansion of the expression but canīt split the expression the same way as in part a

    e.g. a^n=(1+(a-1))^n

    Maybe I donīt need to do the binomial but use what I got i part a somehow?
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  4. #4
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    Quote Originally Posted by wilbursmith View Post
    Ok I think the a part is rather clear now. The only way that I see a^n is bigger than each term though is that a sum is always bigger than any of itīs terms assuming all positive (not much proof there). Is there a more convincing way of seeing that a^n actually is bigger than any of the terms?
    A sum of positive terms is always greater than any of its individual terms. That should be convincing enough for anyone.

    Quote Originally Posted by wilbursmith View Post
    To the b part. I thought that \lim\frac{a^n}{n}\rightarrow\infty since numerator grows faster than denominator, but I quess my prof will repeat my name three times with decreasing pitch while shaking his head as he reads my conclusion.

    I wanted to do a binomial expansion of the expression but canīt split the expression the same way as in part a

    e.g. a^n=(1+(a-1))^n

    Maybe I donīt need to do the binomial but use what I got i part a somehow?
    Use the same binomial expansion as in part (a). But instead of saying that a^n is greater than the term {n\choose1}(a-1), say that a^n is greater than the next term {n\choose2}(a-1)^2.
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  5. #5
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    (b) Show that \lim\frac{a^n}{n}\rightarrow\infty

    Ok, since a^n > than any of the terms as n\rightarrow\infty then

    a^n>\binom{n}{2}(a-1)^2. Divide by (positive) n

    \frac{a^n}{n}>\binom{n}{2}\frac{(a-1)^2}{n}=\frac{n!(a-1)^2}{2n(n-2)!}=\frac{n(n-1)(n-2)...(a-1)^2}{2n(n-2)(n-3)...}=\frac{(n-1)(a-1)^2}{2}\rightarrow\infty as n\rightarrow\infty

    How about this?
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  6. #6
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    Quote Originally Posted by wilbursmith View Post
    (b) Show that \lim\frac{a^n}{n}\rightarrow\infty

    Ok, since a^n > than any of the terms as n\rightarrow\infty then

    a^n>\binom{n}{2}(a-1)^2. Divide by (positive) n

    \frac{a^n}{n}>\binom{n}{2}\frac{(a-1)^2}{n}=\frac{n!(a-1)^2}{2n(n-2)!}=\frac{n(n-1)(n-2)...(a-1)^2}{2n(n-2)(n-3)...}=\frac{(n-1)(a-1)^2}{2}\rightarrow\infty as n\rightarrow\infty

    How about this?
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  7. #7
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    The assignment has a (c) part as well. Hope to get some help on that one too.

    c) Show that for every k\in N

    \lim_{n\rightarrow\infty}n^ka^{-n}=0

    Earlier I stated that a^n>\binom{n}{2}(a-1)^2<br />
and so \frac{1}{a^n}<\frac{1}{\binom{n}{2}(a-1)^2}<br />
multiply by n we get

    \frac{n}{a^n}<\frac{n}{(a-1)^2}\cdot\frac{2(n-2)(n-3)...}{n(n-1)(n-2)(n-3)...}=\frac{2}{(a-1)^2(n-1)}\rightarrow0 as n\rightarrow\infty

    Is this ok? And if so, how do I show that this is true for every k\in N ...induction? How?

    Again it feels natural that if I keep multiplying both sides with (positive)n, k times Iīll still get left side less than right side which tends to zero as n tends to infinity.

    hmm?
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