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Thread: 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 $\displaystyle a>1$ then $\displaystyle \lim_{n\rightarrow\infty}a^n=+\infty$

    Hereīs my brave attempt:

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


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

    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
    $\displaystyle \lim_{n\rightarrow\infty}n^{-1}a^n=+\infty$

    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 $\displaystyle a>1$ then $\displaystyle \lim_{n\rightarrow\infty}a^n=+\infty$

    Hereīs my brave attempt:

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

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

    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 $\displaystyle a^n=1+\binom{n}{1}(a-1)+\binom{n}{2}(a-1)^2+...+\binom{n}{n}(a-1)^n.$ 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 $\displaystyle a^n$ is greater than any single term on the right. In particular, $\displaystyle a^n > \binom{n}{1}(a-1)$, and that tends to infinity as n does.

    Quote Originally Posted by wilbursmith View Post
    The b part:

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

    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 $\displaystyle a^n$ 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 $\displaystyle a^n$ actually is bigger than any of the terms?

    To the b part. I thought that $\displaystyle \lim\frac{a^n}{n}\rightarrow\infty$ 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.$\displaystyle a^n=(1+(a-1))^n$

    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 $\displaystyle a^n$ 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 $\displaystyle a^n$ 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 $\displaystyle \lim\frac{a^n}{n}\rightarrow\infty$ 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.$\displaystyle a^n=(1+(a-1))^n$

    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 $\displaystyle a^n$ is greater than the term $\displaystyle {n\choose1}(a-1)$, say that $\displaystyle a^n$ is greater than the next term $\displaystyle {n\choose2}(a-1)^2$.
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  5. #5
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    (b) Show that $\displaystyle \lim\frac{a^n}{n}\rightarrow\infty$

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

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

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

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

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

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

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

    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 $\displaystyle k\in N $

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

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

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

    Is this ok? And if so, how do I show that this is true for every $\displaystyle 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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