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Converge or Diverges?

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to find whether the sequence converges or diverges
    1.(1+ 1/n)^2n


    2.arctan x


    2. Relevant equations



    3. The attempt at a solution
    1. My answer is wrong but I don't know how I can get the right answer
    My answer
    (1 + 1/n) ^2n becomes 1 because 1/n tends to zero and 2n become infinitely big and
    1 ^ infinity is 1. So the answer is 1 but the answer key says e^2

    2. I don't know how to approach it
     
  2. jcsd
  3. Apr 3, 2012 #2
    1. The limit of (1+ 1/n)^n as n goes towards infinity is e. Since the question asks (1+ 1/n)^2n the answer is e^2.

    Just take the limit of (1+ 1/n)^2n as n goes towards infinity.

    2. the limit as arctan(n) goes to inifinity is infinity, thus arctan(n) diverges.
     
  4. Apr 3, 2012 #3
    [tex]\lim_{n\rightarrow \infty} \left (\left ( 1+\frac{1}{n} \right )^n \right)^2[/tex]
    [tex]\left ( \lim_{n\rightarrow \infty} \left ( 1+\frac{1}{n} \right )^n \right)^2[/tex]

    So let's just ignore the squaring for now. Whatever we find, we will square it.

    [tex] I =\lim_{n\rightarrow \infty} \left ( 1+\frac{1}{n} \right )^n[/tex]

    [tex] ln(I) =\lim_{n\rightarrow \infty} ln\left (\left ( 1+\frac{1}{n} \right )^n \right )[/tex]
    [tex] ln(I) =\lim_{n\rightarrow \infty} nln\left (1+\frac{1}{n} \right )[/tex]

    This is an indeterminate form of zero times infinity. Let
    [tex] N = \frac{1}{n}[/tex]
    then
    [tex] ln(I) =\lim_{n\rightarrow 0} \frac{ln\left (1+N \right )}{N}[/tex]

    Can you take it from here?
     
  5. Apr 3, 2012 #4
    For number 2, have you looked at the graph of arctan x?
     
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