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Telescoping sum

  1. Jan 2, 2015 #1
    1+3+5+...+(2n-1)=∑(2k-1)

    but (2k-1)=k2-(k-1)2
    summing we use the telescoping property and deduce that ∑(2k-1)=n2-02=n2

    This seems accurate to me. Now my question is this a proper use of the telescoping property. In the least it reveals the proper answer, which can then be proved by induction.
     
  2. jcsd
  3. Jan 2, 2015 #2

    Stephen Tashi

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    You example illustrates the use of a telescoping sum to reach a correct conclusion about finite sums. This is a reliable method of reasoning about finite sums. Is that what you mean by the words "accurate" and "proper" ?

    The problem of finding simple expressions for finite sums can be regarded as the problem of "anti-differencing", which has an analogy to antidifferentiation in calculus. You can find material online about doing anti-differencing and there are rules for it that are analogous to those used for antidifferentiation. For example, there is Summation By Parts ( http://en.wikipedia.org/wiki/Summation_by_parts ) which is analgous to Integration By Parts.
     
  4. Jan 2, 2015 #3
    Yes to your first question. I was also interested in how one would determine the limit of the expression Σ[k2-(k-1)2]=Σ(2k-1). How would one know if it were to approach 0 vs. 1.
    I assumed that it went to zero because
    i. (k+1)2=k2+2k+1=(k+1)2-k2=2k+1 then summing over n we have [(n+1)2-(1+n)]/2=∑k. subtracting 1 from the terms in i. gave [k2-(k-1)2]→(n+1-1)2-(1-1)2=n2. I did not know how to determine the limit which is why I was curious if this is a good way to figure out the sum vs. some unknown way. If there is some other way, what would it be?
     
  5. Jan 2, 2015 #4

    Stephen Tashi

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    What limit are you talking about? The limit of the expression as k approaches infinity? - or the limit of the expression as k approaches zero? - or as k approaches some other number?
     
  6. Jan 2, 2015 #5
    Nevermind, I figured out my mistake. I was thinking about it in the wrong way. It was starting with k=1 and ki∈ℕ with k1<k2.
    Thank you for your help.
     
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