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Math Help - What power should I use for a root test over a series?

  1. #1
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    What power should I use for a root test over a series?

    To test for convergence of a series, I would usually take the limit of the series with root test's power 1/n.

    But in some cases, I was told that I could use the power of the coefficient of a power series. For instance, in this series {\textstyle \underset{p=2}{\sum}\frac{(4)^{2p}}{9^{p}p^{5/4}}(\frac{5}{4}-x)^{2p}}, its coefficient has a power of 2p. I could just take the limit of the coefficient, \underset{p\rightarrow\infty}{\lim}\frac{1}{\left[\frac{(4)^{2p}}{9^{p}p^{5/4}}\right]^{1/2p}}=\frac{3}{4}. So 3/4 is the radius of convergence, which is correct. If I took the limit of the root test with power 1/n, the answer would be wrong.

    But then in another series like this one: \underset{n=2}{\sum}\frac{n^{n^{2}}e^{3n}}{(2+n)^{  n^{2}}2^{2n}}4^{2n}(x+\frac{5}{4})^{2n}, when I take the limit of its coefficient with its power 2n, \underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\cdot e^{1/2}}, the answer for the radius of convergence is wrong. The correct answer is 1/(4e). Apparently, I will only get the 1/(4e) as the answer if I took 1/n in my root test instead of 1/2n.

    In this case, how would I know when I should take 1/n or 1/2n or even 1/(2n+1) in my root test for a series?
    Thanks!
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  2. #2
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    Quote Originally Posted by xEnOn View Post
    To test for convergence of a series, I would usually take the limit of the series with root test's power 1/n.

    But in some cases, I was told that I could use the power of the coefficient of a power series. For instance, in this series {\textstyle \underset{p=2}{\sum}\frac{(4)^{2p}}{9^{p}p^{5/4}}(\frac{5}{4}-x)^{2p}}, its coefficient has a power of 2p. I could just take the limit of the coefficient, \underset{p\rightarrow\infty}{\lim}\frac{1}{\left[\frac{(4)^{2p}}{9^{p}p^{5/4}}\right]^{1/2p}}=\frac{3}{4}. So 3/4 is the radius of convergence, which is correct. If I took the limit of the root test with power 1/n, the answer would be wrong.

    But then in another series like this one: \underset{n=2}{\sum}\frac{n^{n^{2}}e^{3n}}{(2+n)^{  n^{2}}2^{2n}}4^{2n}(x+\frac{5}{4})^{2n}, when I take the limit of its coefficient with its power 2n, \underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\cdot e^{1/2}}, the answer for the radius of convergence is wrong. The correct answer is 1/(4e). Apparently, I will only get the 1/(4e) as the answer if I took 1/n in my root test instead of 1/2n.

    In this case, how would I know when I should take 1/n or 1/2n or even 1/(2n+1) in my root test for a series?
    Thanks!

    It is ALL the same! In the first example you use p since that is the running index in the power

    series, and in the second you should use n because that is the index....and if there's an example of a power series

    with running index equal to Gua, then you must use the Gua-root test...

    Tonio
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  3. #3
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    The running index variable is p. But in the first example, I used 1/(2p) as my root test to get the correct answer because I was told that the power of the co-efficient of the series is 2p and I need to use 1/(2p).

    But in the second example, the co-efficient also has 2n as its power but when I use 1/(2n) for the root test, it gives me a wrong answer.

    So it doesn't look quite the same, isn't it?
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  4. #4
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    Quote Originally Posted by xEnOn View Post
    The running index variable is p. But in the first example, I used 1/(2p) as my root test to get the correct answer because I was told that the power of the co-efficient of the series is 2p and I need to use 1/(2p).

    But in the second example, the co-efficient also has 2n as its power but when I use 1/(2n) for the root test, it gives me a wrong answer.

    So it doesn't look quite the same, isn't it?


    I think there's a confusion here: \displaystyle{\lim\limits_{2p\to\infty}\left(\frac  {4^{2p}}{9^pp^{5/4}}\right)^{1/2p}=\frac{4}{\sqrt{9}}\lim\limits_{2p\to\infty}\fr  ac{1}{\left(p^{5/4}\right)^{1/2p}}=\frac{4}{3}}

    and all it's fine.

    About the second series we get what you said is incorrect, but the series itself is a little weird: why

    to write \displaystyle{\frac{4^{2n}}{2^{2n}}\mbox{ instead of simply }=2^{2n} ? Perhaps there's a mistake there?

    I think the radius of the series, as you wrote it, must indeed be \displaystyle{\frac{1}{2\sqrt{e}}} ...

    Tonio
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  5. #5
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    Thanks! Just to clarify a little more. The series converges to \displaystyle{\frac{1}{2\sqrt{e}}} because I took  \frac{1}{2n} as the root in the test, ie: \underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\sqrt{e}}}.

    But if I apply the more generic root in the test,  \frac{1}{n} , then I will get  \frac{1}{4e} as the result, ie: \underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/n}}=\frac{1}{4e}}.

    Then how I know what I need to take as the root in the test? Can I safely say that as long as I take the power of the (x+a), ie: (x+\frac{5}{4})^{2n}, in the test then I will get the correct radius of convergence for the series?
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  6. #6
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    Quote Originally Posted by xEnOn View Post
    Thanks! Just to clarify a little more. The series converges to \displaystyle{\frac{1}{2\sqrt{e}}} because I took  \frac{1}{2n} as the root in the test, ie: \underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\sqrt{e}}}.

    But if I apply the more generic root in the test,  \frac{1}{n} , then I will get  \frac{1}{4e} as the result, ie: \underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/n}}=\frac{1}{4e}}.

    Then how I know what I need to take as the root in the test? Can I safely say that as long as I take the power of the (x+a), ie: (x+\frac{5}{4})^{2n}, in the test then I will get the correct radius of convergence for the series?

    First, it's not the series' but the limit of the inverse of its general term that equals 2e^{-1/2} , second

    I think this is the actual limit since the root test must be taken with respect to the index of the variable, which

    in this case it's x + 5/4 ...

    The above is what I think but perhaps someone else could throw some more light into this.

    Tonio
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