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Math Help - Something wrong with my reasoning?

  1. #1
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    Something wrong with my reasoning?

    I was asked to prove that \frac{c^n}{n!} is a null sequence, that is, lim_{n->\infty}\frac{c^n}{n!}=0. c is a real number.

    I said that the sequence tends to zero if for all n sufficiently large the denominator grows faster than the numerator. If this is true for all n sufficiently large then the denominator is infinitely larger than the numerator in the limit and so the limit is zero.

    The numerator grows at the rate \frac{c^{n+1}}{c^n}=c and the denominator grows at the rate \frac{(n+1)!}{n!}=n+1
    For any c there exists n0 such that n+1>c for all n>n0 therefore the sequence is a null sequence.

    I got zero marks for this, the lecturer later gave the solution in the image I attached. Maybe he was just being picky but I don't see what is wrong with my proof, could someone please show me where I went wrong if I went wrong?
    Attached Thumbnails Attached Thumbnails Something wrong with my reasoning?-sequence-solution.png  
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  2. #2
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    Re: Something wrong with my reasoning?

    Quote Originally Posted by Shakarri View Post
    I was asked to prove that \frac{c^n}{n!} is a null sequence, that is, lim_{n->\infty}\frac{c^n}{n!}=0. c is a real number.
    I said that the sequence tends to zero if for all n sufficiently large the denominator grows faster than the numerator. If this is true for all n sufficiently large then the denominator is infinitely larger than the numerator in the limit and so the limit is zero.
    The numerator grows at the rate \frac{c^{n+1}}{c^n}=c and the denominator grows at the rate \frac{(n+1)!}{n!}=n+1
    For any c there exists n0 such that n+1>c for all n>n0 therefore the sequence is a null sequence.
    I got zero marks for this, the lecturer later gave the solution in the image I attached. Maybe he was just being picky but I don't see what is wrong with my proof, could someone please show me where I went wrong if I went wrong?
    In my view he is being picky. You can show that \sum\limits_{k = n}^\infty  {\frac{{{c^k}}}{{k!}}} converges by the ratio test.
    Thanks from Shakarri and topsquark
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  3. #3
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    Re: Something wrong with my reasoning?

    Thank you for the reassurance. Something similar came up on the exam today! I hope he's in a better mood this time
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    Re: Something wrong with my reasoning?

    Quote Originally Posted by Shakarri View Post
    I don't see what is wrong with my proof, could someone please show me where I went wrong if I went wrong?
    You can find this out yourself if you write down your proof in more detail. If you really understand it, you should be able to fill in as much detail as necessary, up to axioms of real numbers.

    I see the following problematic places in your proof.

    Quote Originally Posted by Shakarri View Post
    I said that the sequence tends to zero if for all n sufficiently large the denominator grows faster than the numerator.
    I am not sure the phrase "f(n) grows faster than g(n)" has a universally accepted definition. Besides, if it has a reasonable definition, there is probably no need to say "for all n sufficiently large".

    Quote Originally Posted by Shakarri View Post
    The numerator grows at the rate \frac{c^{n+1}}{c^n}=c and the denominator grows at the rate \frac{(n+1)!}{n!}=n+1
    Again, unless your course formally defined the phrase "f(n) grows at the rate g(n)", it is not good to use it.

    Quote Originally Posted by Shakarri View Post
    For any c there exists n0 such that n+1>c for all n>n0 therefore the sequence is a null sequence.
    This seems like a huge jump from a trivial premise to a nontrivial conclusion.
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