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Math Help - exponential limit

  1. #1
    Junior Member winterwyrm's Avatar
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    exponential limit

    ok, the lim of the nth root of (1+3n) Thanks in advance!
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  2. #2
    Eater of Worlds
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    What does n approach?.

    I assume you mean \lim_{n\to {\infty}}\left(1+3n\right)^{\frac{1}{n}}

    If so, let t=\frac{1}{n}

    Then, \lim_{t\to 0}\left(1+\frac{3}{t}\right)^{t}=1
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by winterwyrm View Post
    ok, the lim of the nth root of (1+3n) Thanks in advance!
    By the connection between the root and ratio test

    \lim_{n\to\infty}(1+3n)^{\frac{1}{n}}=\lim_{n\to\i  nfty}\frac{1+(3n+1)}{1+3n}=1
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    By the connection between the root and ratio test

    \lim_{n\to\infty}(1+3n)^{\frac{1}{n}}=\lim_{n\to\i  nfty}\frac{1+(3n+1)}{1+3n}=1
    Que est?
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  5. #5
    Moo
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    Quote Originally Posted by mr fantastic View Post
    Que est?
    "Qu'est-ce que c'est ?" ^^

    If you want to know, you can remember that when dealing with power series and looking for the radius of convergence, you can choose between the ratio or the power test. Both will give the same result

    Now, if you want to prove it, that's another matter lol
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Que est?
    Qu-est que tu veux?
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  7. #7
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    Quote Originally Posted by Moo View Post
    "Qu'est-ce que c'est ?" ^^

    If you want to know, you can remember that when dealing with power series and looking for the radius of convergence, you can choose between the ratio or the power test. Both will give the same result

    Now, if you want to prove it, that's another matter lol
    But where is the power series here? Is it meant to be \sum (1 + 3n) x^n .....?

    Quote Originally Posted by Mathstud28
    By the connection between the root and ratio test


    If a_n = 1 + 3n then the ratio is wrong for a start.

    By the way, regarding "By the connection between the root and ratio test" ...... What happens when you apply the ratio test and the nth root test to \sum_{n=1}^{\infty} 2^{(-1)^n - n} .....?

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  8. #8
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    Quote Originally Posted by Moo View Post
    Now, if you want to prove it, that's another matter lol
    The proof is not that bad.
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  9. #9
    Moo
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    Quote Originally Posted by mr fantastic View Post
    But where is the power series here? Is it meant to be \sum (1 + 3n) x^n .....?
    Nah, I was talking about when you want to find the radius of convergence of a series, you can use either the ratio test either the root test to reach the same result.
    This is a way to "remember" what mathstud used

    @ TPH : would you mind posting it ? ^^
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  10. #10
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    Quote Originally Posted by Moo View Post
    Nah, I was talking about when you want to find the radius of convergence of a series, you can use either the ratio test either the root test to reach the same result.
    This is a way to "remember" what mathstud used

    @ TPH : would you mind posting it ? ^^
    I understand what you are saying and fully agree (I'm not that senile - actually I knew this already. Hmmmppph .... learned it no more than 80 or 90 years ago).

    But what Mathstud said is wrong in the context he used it in. (Otherwise I have completely misunderstood).
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I understand what you are saying and fully agree (I'm not that senile - actually I knew this already. Hmmmppph .... learned it no more than 80 or 90 years ago).

    But what Mathstud said is wrong in the context he used it in. (Otherwise I have completely misunderstood).
    Well, Mr. F, in what context is that?
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  12. #12
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    Quote Originally Posted by Mathstud28 View Post
    Well, Mr. F, in what context is that?
    Quote Originally Posted by Mathstud28
    By the connection between the root and ratio test
    *Ahem*

    Quote Originally Posted by Mr Fantastic
    [snip]
    What happens when you apply the ratio test and the nth root test to \sum_{n=1}^{\infty} 2^{(-1)^n - n} .....?
    Since one test gives 1/2 and the other gives 1, how do you justify your approach to the limit posed by the OP ...... ?
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    *Ahem*



    Since one test gives 1/2 and the other gives 1, how do you justify your approach to the limit posed by the OP ...... ?
    Are you really going to be that semantical? Well to answer your question, I never said every limit can be done this way, what you provided is a case where the root test and the ratio test differ. What I am talking about is much more specific, it is

    By the connection

    \lim_{n\to\infty}f(n)^{\frac{1}{n}}=\lim_{n\to\inf  ty}\frac{f(n+1)}{f(n)}

    And f(n) cannot have a variable exponent. I will post back later with a more in-depth reasoning.
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  14. #14
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    Quote Originally Posted by Mathstud28 View Post
    Are you really going to be that semantical? Well to answer your question, I never said every limit can be done this way, what you provided is a case where the root test and the ratio test differ. What I am talking about is much more specific, it is

    By the connection

    \lim_{n\to\infty}f(n)^{\frac{1}{n}}=\lim_{n\to\inf  ty}\frac{f(n+1)}{f(n)}

    And f(n) cannot have a variable exponent. I will post back later with a more in-depth reasoning.
    No need. Your reply has provided sufficient clarification to your earlier post.
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