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Math Help - Simple proof of integration formula from geometric series

  1. #1
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    Post Simple proof of integration formula from geometric series

    The last 2 lines of the proof for the formula for integration of x^a, for positive a, not the second half which deals with negative alpha. I don't understand how he gets that sum from (q-1)/(q^(1+a)-1)). And then I don't know how he gets the final result 1/(1+a). The attached file is from page 133 of Courant and John's Intro to analysis and calculus.
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  2. #2
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    Presumably you know that the sum of a geometric series \sum_{n= 1}^\alpha ar^n= \frac{a(1- r^{\alpha+ 1}{1- r}. With a= 1, r= q, that is \sum_{n=1}^{\alpha} q^n= \frac{q^{\alpha+1}- 1}{q- 1}.

    Taking the reciprocal of both sides gives
    \frac{1}{\sum_{n=1}^{\alpha} q^n}= \frac{q-1}{q^{\alpha+1}- 1}
    the next to the last formula. Notice that there are \alpha+ 1 terms in that sum. As the book says, as n goes to infinity, all those powers of q go to 1 so we are adding 1 \alpha+ 1 times- the sum is \alpha+ 1 so the left side is \frac{1}{\alpha+1}.

    Putting that into the fourth formula on the page,
    F_n= (b^{\alpha+ 1}- a^{\alpha+1})q^\alpha\frac{q- 1}{q^{\alpha+1}- 1}
    gives the final result.
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