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Math Help - Limits

  1. #1
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    Limits

    Determine the following limits if they exists. If the limit exists, establish the convergence.

    (a) \lim _{n\rightarrow\infty} (\sqrt{(n+a)(n+b)} - n where a,b > 0.

    (b) \lim _{n\rightarrow\infty} (n!)^{\frac{1}{n^2}}
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by yusukered07 View Post
    Determine the following limits if they exists. If the limit exists, establish the convergence.

    (a) \lim _{n\rightarrow\infty} (\sqrt{(n+a)(n+b)} - n where a,b > 0.

    (b) \lim _{n\rightarrow\infty} (n!)^{\frac{1}{n^2}}
    Let L=\lim_{n\to\infty}\left(n!\right)^{\frac{1}{n^2}}  =\lim_{n\to\infty}\left(\left(n!\right)^{\frac{1}{  n}}\right)^{\frac{1}{n}}, applying the connection between the root and ratio tests we see that this is equal to L=\lim_{n\to\infty}\frac{((n+1)!)^{\frac{1}{n+1}}}  {(n!)^{\frac{1}{n}}}=\lim_{n\to\infty}\frac{(n!)^{  \frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}\cdot (n+1)^{\frac{1}{n+1}}=\lim_{n\to\infty}(n!)^{\frac  {-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{  1}{n^2}}}=\frac{1}{L}. Thus, L=\pm 1 but since it is clearly positive we may conclude that L=1.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Is...

    \sqrt{(n+a)\cdot (n+b)} - n = \frac{(n+a)\cdot (n+b)}{\sqrt{(n+a)\cdot (n+b)} + n} =

    = \frac{n\cdot (a+b)+ a b }{\sqrt{(n+a)\cdot (n+b)} + n} = \frac{a + b + \frac{a b}{n}}{1+ \sqrt{1 + \frac{a+b}{n} + \frac{a b}{n^{2}}}} (1)

    ... so that is...

    \lim_{n \rightarrow \infty} \sqrt{(n+a)\cdot (n+b)} - n = \frac{a+b}{2} (2)

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    Let L=\lim_{n\to\infty}\left(n!\right)^{\frac{1}{n^2}}  =\lim_{n\to\infty}\left(\left(n!\right)^{\frac{1}{  n}}\right)^{\frac{1}{n}}, applying the connection between the root and ratio tests we see that this is equal to L=\lim_{n\to\infty}\frac{((n+1)!)^{\frac{1}{n+1}}}  {(n!)^{\frac{1}{n}}}=\lim_{n\to\infty}\frac{(n!)^{  \frac{1}{n+1}}}{(n!)^{\frac{1}{n}}}\cdot (n+1)^{\frac{1}{n+1}}=\lim_{n\to\infty}(n!)^{\frac  {-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{  1}{n^2}}}=\frac{1}{L}. Thus, L=\pm 1 but since it is clearly positive we may conclude that L=1.
    I am not sure this is fully rigorous: for instance, don't you need to prove first that the limit exists? (and a few steps would need justifications, like \lim_{n\to\infty}(n!)^{\frac{-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{  1}{n^2}}} or the previous one)

    I would rather say: readily, 1\leq n!=n(n-1)\cdots 1\leq n^n, hence 1\leq (n!)^{\frac{1}{n^2}}=\exp(\frac{\log (n!)}{n^2})\leq \exp(\frac{\log n}{n}). Since \frac{\log n}{n}\to 0 (and \exp is continuous at 0), the limit indeed exists and equals 1.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Laurent View Post
    I am not sure this is fully rigorous: for instance, don't you need to prove first that the limit exists? (and a few steps would need justifications, like \lim_{n\to\infty}(n!)^{\frac{-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{  1}{n^2}}} or the previous one)

    I would rather say: readily, 1\leq n!=n(n-1)\cdots 1\leq n^n, hence 1\leq (n!)^{\frac{1}{n^2}}=\exp(\frac{\log (n!)}{n^2})\leq \exp(\frac{\log n}{n}). Since \frac{\log n}{n}\to 0 (and \exp is continuous at 0), the limit indeed exists and equals 1.
    I don't think that the point of the exercise was to be fully rigorous. Besides, I don't think I made any extraordinary leaps.
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    I don't think that the point of the exercise was to be fully rigorous. Besides, I don't think I made any extraordinary leaps.
    I was actually a bit euphemistic in my answer : assuming the existence of the limit in the first place and playing with it is really really not a good habit to take (no harm intended!). Especially when, like here, we are asked "if the limit exists, establish the convergence"; but certainly not only in this case.
    It is not because the only "possible" limit is 1 that the sequence converges to 1.
    I'm just saying...
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  7. #7
    Moo
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    Quote Originally Posted by Drexel28 View Post
    I don't think that the point of the exercise was to be fully rigorous. Besides, I don't think I made any extraordinary leaps.
    If you act this way when you are a mathematician, you'll be in some kind of trouble
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Moo View Post
    If you act this way when you are a mathematician, you'll be in some kind of trouble
    Haha! I agree

    That said...I struggle with complete rigor and not giving the OP the full answer. My response was meant to be a nudge for which they would fill in the details
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