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Ratio Test/ Simple English

  1. Nov 22, 2004 #1
    Can anyone explain what exactly the ratio test is comparing?

    undefinedSome say it is being compared to a geometric series...I love geo. series and I don't see how the test involves them.

    Thanx,
    Erin
     
  2. jcsd
  3. Nov 23, 2004 #2

    Galileo

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    Ok, I`m a bit rusty onseries, but let's see if I remember why the ratio test works.

    Ratio test:

    If [itex]\limsup \left| \frac{a_{n+1}}{a_n}\right| <1[/itex] then the series [itex]\sum a_n[/itex] converges absolutely.
    If [itex]\limsup \left| \frac{a_{n+1}}{a_n}\right| >1[/itex] then the series [itex]\sum a_n[/itex] diverges.

    I`ll just take the case where the limit < 1. And see why it converges absolutely in that case.
    Suppose [itex]\limsup \left| \frac{a_{n+1}}{a_n}\right| =\rho<1[/itex]
    Then for all [itex]\epsilon >0[/itex], there is an N, so that for n>N:
    [tex]\left| \frac{a_{n+1}}{a_n}\right-\rho|<\epsilon[/tex]
    or
    [tex]-\epsilon + \rho < \frac{a_{n+1}}{a_n} < \epsilon +\rho[/tex]
    Since [itex]\rho[/itex] is smaller than 1, we can choose an epsilon small enough so that [itex]\epsilon +\rho=r<1[/itex]. Then if N is big enough:
    [tex]|\frac{a_{n+1}}{a_n} |< r[/tex]
    or
    [tex]|a_{n+1}| < |a_n|r[/tex]
    and also
    [tex]|a_{n+2}| < |a_{n+1}|r<|a_n|r^2[/tex]
    and so on...
    [tex]|a_{n+k}| < |a_n|r^k[/tex]
    for n>N.
    Therefore:
    [tex]\sum_{n=0}^{\infty}|a_n|\leq \sum_{n=N}^{\infty}|a_n|<\sum_{n=0}^{\infty}|a_N|r^n[/tex]
    So the series is smaller than a convergent geometric series.
     
    Last edited: Nov 23, 2004
  4. Nov 23, 2004 #3

    mathwonk

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    same thing in words,

    tests of convergence are usually comparisons. i.e. if a series of positive terms is smaller term by term than some convergent series, then it is also convergent.

    This is true even if it only holds from some finite point onward.

    The best series to compare with is the geometric series. A geometric series is a series in which the ratio of each term divided by the previous term is the same number. if that number is less than one it converges.

    hence if a series has the property that the ratios of each term divided by the previous term eventually all become less than some number r with 0<r<1, then that original series is eventually smaller than the geometric series with ratio r, hence also convergent.

    the best possible treatment of convergence of series is in richard courants calculus book.
     
  5. Nov 27, 2004 #4
    Thanks...So may I assume that the ratio of successive terms of a power series ,
    _
    \
    / cn(x-a)^n, near infinity is constant? Because when we do the ratio test, we are basically testing the ratio of successive terms at infinity.
    _
     
  6. Nov 28, 2004 #5

    mathwonk

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    you do not need them to be constant, just for all of them eventually i.e. near infinity) to be less than some number which itself is less than one. the easy case is where these ratios converge to a limit less than one.

    bewst npossible example: exponential series: the nth term is

    x^n/n! so the ratio of the nth by the n-1st term is x/n. as n goes to infinity this approaches 0, no matter what x is. so it converges no matter what x is.
     
    Last edited: Nov 28, 2004
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