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Question about one part of the Ratio Test proof

  1. Mar 15, 2012 #1
    Hi,

    there is a proof of the ratio test that I have seen a couple of times here:

    http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx


    I dont know how to use latex, so abs(x) will mean absolute value

    I am ok with this:

    abs( aN+1) < r abs(aN)
    I dont understand where the r2 come from in the next line
    abs( aN+2) < r abs(aN+1)< r2 abs(aN)

    Similarly, I dont understand how to get to the generalization of
    abs( aN+k) < rk abs(aN+1)


    I should probably add that I understand all other aspects (i.e. using comparison test), but since I don't understand this part, my understanding of the proof fails.

    Thanks,

    Ivan77
     
    Last edited: Mar 15, 2012
  2. jcsd
  3. Mar 15, 2012 #2
    Just in case your were wondering, you don't need LaTeX for absolute value, the key is right above your Enter key on your keyboard (PC): |

    So we know [itex]\left|\frac{a_{n+1}}{a_n}\right| < r[/itex] which is the same as [itex]|a_{n+1}| < r |a_n|[/itex]. This is saying that the n+1st term (the term after the nth term) in the sequence is the nth term multiplied by r.

    What they are doing in the proof is going backwards. Think of it this way, it might help you better visualize it:

    Starting with [itex]|a_{N+1}| < r |a_N|[/itex] and [itex]|a_{N+2}| < r |a_{N+1}|[/itex] we can substitute for [itex]|a_{N+1}|[/itex] below:

    [itex]
    \begin{eqnarray*}
    |a_{N+2}| &<& r |a_{N+1}|\\
    &<& r (r |a_{N}|) = r^2 |a_N|
    \end{eqnarray*}
    [/itex]

    And you can continue this process

    [itex]
    \begin{eqnarray*}
    |a_{N+3}| &<& r |a_{N+2}|\\
    &<& r (r^2 |a_{N}|) = r^3 |a_N|
    \end{eqnarray*}
    [/itex]


    Does that help or is it still confusing?
     
  4. Mar 15, 2012 #3
    Hugely appreciate the reply scurty! It answered my question. I've been staring at this proof for a couple of hours.

    I was getting stuck on the logic of how we would know that the r^2 would not make the term less than the An+1 rather that using the inequality given.

    This is what I was gettign stuck on:

    r|aN+1| < r(r|aN|)

    has to be true since

    since
    |aN+1| < r|aN|

    and multiplying by a factor of r will not change that fact.

    Why couldn't I see that simply distributing the r in the right term would answer the question?
    The odd thing about my self studying calculus is that the concepts as well answering questions/application is not a problem, but now and then, I get totally stuck on what usually ends up being a minor point.

    Any tips?
     
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