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Limit of series

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data
    why we need to make x as absolute value ? as we can see, the original is x , why we we need to make x as absolute value ? is the working wrong ?

    2. Relevant equations


    3. The attempt at a solution
     

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  2. jcsd
  3. Feb 2, 2016 #2

    Simon Bridge

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    Look one line up from where you put the circle to see where it comes from.
    What happens to the limit when x is a negative number, and you don't have the absolute value?
     
  4. Feb 2, 2016 #3
    Ya, it come from the top. But, why would we need to put iy as absolute value?
     
  5. Feb 2, 2016 #4

    Simon Bridge

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    Because if you don't the possibility that x may be negative will give you the wrong limit. Try it and see:
    What happens to the limit when x is a negative number, and you don't have the absolute value?
     
  6. Feb 2, 2016 #5

    PeroK

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    It's using the ratio test. This looks at ##\lim_{k \rightarrow \infty} | \frac{a_{k+1}}{a_k}|##

    The modulus should be introduced at the start, not arbitrarily half way through the calculation.
     
  7. Feb 2, 2016 #6
    taking x=-4 , k =2 , the term is -16/27
    taking x=4 , k=2 the term is 16/27 , i couldn't find out why there is a need to put modulus ???
     
  8. Feb 2, 2016 #7

    PeroK

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  9. Feb 2, 2016 #8

    Simon Bridge

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    It's not what happens to the term, it's what happens to the limit.
    PeroK has a good link (thanks).
     
  10. Feb 2, 2016 #9
    what will happen to the limit ? i really have no idea...
     
  11. Feb 2, 2016 #10
    well, for this part of the notes, why do we need to make r is between L and 1 , and we take r is larger than L at the lower part .... isn't the L itself is the factor r ?
     

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  12. Feb 2, 2016 #11

    Samy_A

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    No , L doesn't necessarily satisfy ##| \frac{a_{k+1}}{a_k}| \leq L## for k sufficiently large.
    Any r with ##L<r<1## does satisfy ##| \frac{a_{k+1}}{a_k}| \leq r## for k sufficiently large.

    As an example, take ##0<L<1##, and define ##(a_n)_n## by:
    ##a_1=1,\ a_{n+1}=a_n(L+\frac{1}{n})##

    Then ##|\frac{a_{n+1}}{a_n}|=L+\frac{1}{n} > L##, but ##\displaystyle \lim_{n\rightarrow +\infty} |\frac{a_{n+1}}{a_n}| =L##.
     
    Last edited: Feb 2, 2016
  13. Feb 2, 2016 #12

    Ray Vickson

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    If ##L = \lim_{n \to \infty} |a_{n+1}/a_n|## then, for any (small) ##\epsilon > 0## we have ##(L-\epsilon) |a_n| < |a_{n+1}| < (L+\epsilon) |a_n|## for all ##n## sufficiently large; essentially, that is the definition of the limit above. We simply cannot conclude that ##|a_{n+1}| \leq L |a_n|##, and in fact, you can construct examples where this is false.
     
  14. Feb 2, 2016 #13
    can you give example for the explaination above ?
     
  15. Feb 3, 2016 #14

    Samy_A

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    The example in post 11 illustrates ##|\frac{a_{n+1}}{a_n}|>L## for each n, while ##\displaystyle \lim_{n\rightarrow +\infty} |\frac{a_{n+1}}{a_n}| =L##.
     
  16. Feb 4, 2016 #15

    vela

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    Maybe a picture will help. When you say ##\lim_{n\to\infty}\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = L##, it means when ##n## is sufficiently large, ##\left\lvert \frac{a_{n+1}}{a_n} \right\rvert## will be within ##\varepsilon## of ##L##, where we can choose ##\varepsilon>0## arbitrarily. In the picture, that means the terms of the sequence will lie in the green part of the number line. While the terms are close to ##L##, they could be greater than or less than ##L##, so the limit doesn't guarantee that ##\left\lvert \frac{a_{n+1}}{a_n} \right\rvert<L##. However, since ##r>L##, there's some room between ##r## and ##L##, so we can choose ##\varepsilon## such that we can guarantee that ##\left\lvert \frac{a_{n+1}}{a_n} \right\rvert < r## for ##n## sufficiently large.

    limit.png
     
  17. Feb 4, 2016 #16
    what is the purpose of not making L =r , but make L less than r ? isn't the L is the factor = r ?
     
  18. Feb 4, 2016 #17

    vela

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    A proof is an argument, so you're essentially asking, why don't you argue it this way instead. So why don't you try finishing the proof with ##r## set equal to ##L##? You'll see it won't work.
     
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