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Cannot follow this argument pertaining to geometric series.

  1. Jul 27, 2015 #1
    In order to construct a geometric series we do the following:

    chose a number a such that a does not equal 0 and a second number r that is between (-1,1). We call r the ration because it is the ratio, the progression of each term to its predecessor.

    We have An=a+ar+ar^2+ar^3...ar^n

    We multiply An by r. rAn=ar+ar^2+ar^3+ar^4...ar^n+1.

    We subtract rAn from An and we end with the terms: An-rAn=a-ar^n+1


    Here is were i cannot follow the argument. In the book we have An-rAn=a(1-r^n+1)/(1-r).

    Where did the 1-r in the denominator come from? Is it the same idea we invoke to compare two physical or more physical quantities, i.e., ratios of sides of triangles?

    ignoring my lack of understanding of this step. I can see why, s equals the limit as n->infinity of An-rAn

    is a/(1-r). we can pick a value for r between -1 and 1, say 0.5 and raise it to the 1000000 power which will make it go to zero.
  2. jcsd
  3. Jul 27, 2015 #2


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    So on the right they pull out the a: ##\quad a-ar^{n+1} = a\left ( 1-r^{n+1} \right )##

    On the left they do the same! : ##\quad A_n-rA_n = A_n (1-r)##

    So I suspect the book has An = a(1-r^n+1)/(1-r) and not An-rAn = a(1-r^n+1)/(1-r) as you quote (using copy/paste ?)
  4. Jul 29, 2015 #3


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    Actually, r is called the "common ratio".

    Perhaps it would help you to look at the sum "written out".
    [tex]\sum_{n= 0}^\infty ar^n= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot[/tex]

    Write [itex]S= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot[/itex]. Then, subtracting a from both sides, [itex]S- a= ar+ ar^2+ ar^3+ \cdot\cdot\cot= r(a= ar+ ar^2+ ar^3+ \cdot\cdot\cdot[/itex]

    Notice that, because this is an infinite series, we never "lose" the end term so that last sum is just "S" again:
    [itex]S- a= rS[/itex] so that [itex]a= S- rS= (1- r)S[/itex] and [itex]S= \frac{a}{1- r}[/itex].

    A more "rigorous" proof would be to look at the finite sums (because the sum of an infinite series is defined as the limit of its "partial sums").

    [itex]\sum_{n= 0}^N ar^n= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot+ ar^{N-1}+ ar^N[/itex]

    Write [itex]S= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot+ ar^{N-1}+ ar^N[/itex].

    Now do the same as before: [itex]S- a= ar+ ar^2+ar^3+ \cdot\cdot\cdot+ ar^{N-1}+ ar^N[/itex]
    Factor "r" out of the right hand side: [itex]S- a= r(a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^{N- 2}+ ar^{N-1})[/itex]
    Now, because this is a finite sum, there is a "last term" and it is no longer the same as before. To get back the same sum, add [itex]ar^{N+ 1}[/itex] to both sides:
    [tex]S- a+ ar^{N+ 1}= r(a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^{N- 2}+ ar^{N-1})+ ar^{N+ 1}= r(a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^{N- 2}+ ar^{N-1}+ ar^N)[/tex]
    Now that right side is exactly r times the original sum:
    [tex]S- a+ ar^{N+1}= rS[/tex]
    so that
    [tex]S- rS= S(1- r)= a- ar^{N+1}= a(1- r^{N+ 1})[/tex]
    [tex]S= \frac{a(1- r^{N+1})}{1- r}[/tex]

    You can check that this is correct for small values of N. For example for n= 3 the sum is [itex]a+ ar+ ar^2+ ar^3= a(1+ r+ r^2+ r^3)[/itex]. But it is easy to see that [itex](1+ r+ r^2+ r^3)(1- r)= 1- r^4[/itex] so [itex]1+ r+ r^2+ r^3= \frac{1- r^4}{1- r}[/itex].

    And, of course, since -1< r< 1 (an important requirement), as N goes to infinity [itex]r^{N+1}[/itex] goes to 0.
  5. Jul 29, 2015 #4
    Thanks Ivy. The more rigorous proof allowed me to understand completely instantly.

    The topic of series and sequences is very interesting. Do I have to wait for higher mathematics, such as, algebra, number theory, or maybe analysis to learn more?

    I feel that my current calculus books leave out alot of details.

    Is there a book accessible at my level. I have done calculus 1 and 2, linear algebra.

    Or do I need more mathematical knowledge to pursue this idea further?
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