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Math Help - convergent sequences

  1. #1
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    convergent sequences

    Hi,I'm having problem determining whether this sequence is convergent or divergent:
    an= 3n+2n!/1+n^3

    Could someone please show me how to solve this problem. Thankyou very much xx
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  2. #2
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    a_n=\frac{3n+2n!}{1+n^3}

    First the intuition: the sequence n!,\, n\geq 0 grows way quicker than any polynomial. Hence the sequence diverges to +\infty.

    Now for the proof. Because of the intuition, we try to bound a_n from below, keeping the n!: for instance, for n\geq 1, a_n\geq\frac{2n!}{2n^3}=\frac{n!}{n^3} (since 1+n^3 \leq 2n^3).

    In a short while, you'll just have to say " n^3 is negligible compared to n!, hence \frac{n!}{n^3}\to_n+\infty." But at that time, you may prefer a proof of this fact (it depends of what you know already).

    So let b_n=\frac{n!}{n^3}. We prove that (b_n)_n is larger than a divergent geometric sequence. For n\geq 1, \frac{b_{n+1}}{b_n}=\frac{(n+1)!}{(n+1)^3}\frac{n^  3}{n!}=(n+1)\left(\frac{1}{1+\frac{1}{n}}\right)^3. Because \frac{1}{n}\to_n 0, we have thus \frac{b_{n+1}}{b_n}\to_n +\infty. Hence there exists n_0\geq 1 such that, for all n\geq n_0, \frac{b_{n+1}}{b_n}\geq 2, i.e. b_{n+1}\geq 2 b_n. Hence, for n\geq n_0, b_n\geq 2b_{n-1}\geq 2^2 b_{n-2}\geq\cdots\geq 2^{n-n_0}b_{n_0}. Now you know that 2^n\to_n+\infty and b_n\geq 2^n 2^{-n_0}b_{n_0} (and b_{n_0}\neq 0), so that b_n\to_n+\infty.
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  3. #3
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    a_n=\frac{3n+2n!}{1+n^3}
    For n\geq 3,
    \frac{3n+2n!}{n^3+1} \geq \frac{n!}{n^3} \geq \frac{n!}{n(n-1)(n-2)} = (n-3)(n-4)...(2)(1)

    Of course this is divergent.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    For n\geq 3,
    \frac{3n+2n!}{n^3+1} \geq \frac{n!}{n^3} \geq \frac{n!}{n(n-1)(n-2)} = (n-3)(n-4)...(2)(1)
    Yes, thank you for this short solution (what I wrote could have aimed at proving n^k=o_n(n!) for general k). There is a little mistake though, that I'd correct into:
    \frac{n!}{n^3}=\frac{n(n-1)(n-2)}{n^3}(n-3)!=\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)(n-3)!\geq\frac{2}{3}\frac{1}{3}(n-3)!=\frac{2}{9}(n-3)!
    for n\geq 3.
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