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Determining if the series converges

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data
    n=1∑[itex]\sqrt[n]{3}[/itex]*[itex]\sqrt[n]{n}[/itex]



    2. Relevant equations



    3. The attempt at a solution
    I'm thinking that for this series I should use the nth term divergence test.
    I know that the limit of each of them is 1, and the whole series converges to 1, but I can't remember why the limits are 1. Could someone please remind me? :)
     
  2. jcsd
  3. Mar 13, 2014 #2

    jbunniii

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    Your typesetting is messed up. Is this the series you are evaluating?

    $$\sum_{n=1}^{\infty} \sqrt[n]{3} \sqrt[n]{n}$$

    If so, you are correct that ##\lim_{n\rightarrow \infty} \sqrt[n]{3} = 1## and ##\lim_{n\rightarrow \infty} \sqrt[n]{n} = 1##.

    As you said, this implies that ##\lim_{n\rightarrow \infty} \sqrt[n]{3} \sqrt[n]{n} = 1## (the limit of a product is equal to the product of the limits - it's a good idea to check the proof of this if you aren't sure).

    What does that imply about the limit of the series?
     
  4. Mar 13, 2014 #3
    Sorry,that is what I meant! Since the limit doesn't approach 0 it diverges? But why does the limit equal 1 when you plug in infinity?
     
  5. Mar 13, 2014 #4

    jbunniii

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    Regarding why ##\lim_{n\rightarrow \infty} \sqrt[n]{3} = 1##:

    First note that ##\sqrt[n]{3} = 3^{1/n}##. Consider the log of this expression: ##\log(3^{1/n})##. Can you find the limit the log as ##n \rightarrow \infty##? Can you use the result to find the original limit?
     
  6. Mar 13, 2014 #5

    jbunniii

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    Yes, that's right. The terms do not approach zero, so the series diverges.
    See my previous post for a suggestion for ##\lim_{n \rightarrow 1} \sqrt[n]{3}##. The other limit, ##\lim_{n \rightarrow 1} \sqrt[n]{n}##, can be handled similarly but with a slight complication.
     
  7. Mar 13, 2014 #6
    I'm sorry, I still don't quite understand what you mean. I get how you're rewriting the square root, but I got a little lost with the logs.
     
  8. Mar 13, 2014 #7

    jbunniii

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    I claim that ##\lim_{n \rightarrow \infty} \log a_n = \log \lim_{n \rightarrow \infty} a_n## for any convergent sequence of positive values ##a_n##. This is because ##\log## is a continuous function. Let's assume we know this result for now - if you're not sure why it's true, we can come back to it later.

    So in your case, ##a_n = \sqrt[n]{3}##, and we have what I will call Equation 1:
    $$\lim_{n \rightarrow \infty} \log \sqrt[n]{3} = \log \lim_{n \rightarrow \infty} \sqrt[n]{3}$$
    Working with the left hand side, we have ##\log \sqrt[n]{3} = \log (3^{1/n}) = \frac{1}{n} \log 3## by the basic property of logarithms that ##\log x^y = y \log x##.

    Now take the limit of that expression: ##\lim_{n \rightarrow \infty} \frac{1}{n} \log{3} = ???## You should be able to get a numerical answer. Once you have that answer, do you see how Equation 1 allows you to find ##\lim_{n \rightarrow \infty} \sqrt[n]{3}##?
     
  9. Mar 13, 2014 #8
    Okay!! Thanks, I think I get it! :)
     
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