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Math Help - Series

  1. #1
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    Series

    Hi guys, please help with this QNS:

    By considering 1/(1 + a^n-1) - 1/(1 + a^n), deduce Σ (a^n)/[(1 + a^n-1)(1 + a^n)] upper limit: N, lower limit: n = 1


    So this is how I did it:
    1/(1 + a^n-1) - 1/(1 + a^n) = a^n/[(1 + a^n-1)(1 + a^n)] - a^n-1/[(1 + a^n-1)(1 + a^n)]
    (a^n)/[(1 + a^n-1)(1 + a^n)] = 1/(1 + a^n-1) - 1/(1 + a^n) + a^n-1/[(1 + a^n-1)(1 + a^n)]

    => Σ (a^n)/[(1 + a^n-1)(1 + a^n)] = Σ (1/(1 + a^n-1) - 1/(1 + a^n) + a^n-1/[(1 + a^n-1)(1 + a^n)] )

    (Then I use the method of difference by substituting n = 1,2,3.. but I couldn't solve since there isn't a standard way of cancelling the terms)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Blizzardy View Post
    Hi guys, please help with this QNS:

    By considering 1/(1 + a^n-1) - 1/(1 + a^n), deduce Σ (a^n)/[(1 + a^n-1)(1 + a^n)] upper limit: N, lower limit: n = 1


    So this is how I did it:
    1/(1 + a^n-1) - 1/(1 + a^n) = a^n/[(1 + a^n-1)(1 + a^n)] - a^n-1/[(1 + a^n-1)(1 + a^n)]
    (a^n)/[(1 + a^n-1)(1 + a^n)] = 1/(1 + a^n-1) - 1/(1 + a^n) + a^n-1/[(1 + a^n-1)(1 + a^n)]

    => Σ (a^n)/[(1 + a^n-1)(1 + a^n)] = Σ (1/(1 + a^n-1) - 1/(1 + a^n) + a^n-1/[(1 + a^n-1)(1 + a^n)] )

    (Then I use the method of difference by substituting n = 1,2,3.. but I couldn't solve since there isn't a standard way of cancelling the terms)
    Google "telescoping series"

    CB
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  3. #3
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    I'll use k and n for simplicity, and in the second line, we replace k with k+1 for the sum on the left.
    \begin{aligned}& S =  \sum_{1 \le k \le n}\frac{a^k}{(1+a^k)(1+a^{k-1})}  = \frac{a}{a-1}\sum_{1 \le k \le n}\frac{1}{1+a^{k-1}}-\frac{a}{a-1}\sum_{1 \le k \le n}\frac{1}{1+a^k} \\& = \frac{a}{a-1}\sum_{1 \le k+1 \le n}\frac{1}{1+a^{k}}-\frac{a}{a-1}\sum_{1 \le k \le n}\frac{1}{1+a^k} = \frac{a}{a-1}\sum_{0 \le k \le n-1}\frac{1}{1+a^{k}}-\frac{a}{a-1}\sum_{1 \le k \le n}\frac{1}{1+a^k} \\& = \frac{a}{a-1}\left(\frac{1}{1+a^0}\right)-\frac{a}{a-1}\left(\frac{1}{1+a^n}\right)+\frac{a}{a-1}\sum_{1 \le k \le n}\frac{1}{1+a^k}-\frac{a}{a-1}\sum_{1 \le k \le n}\frac{1}{1+a^k} \\& = \frac{a}{a-1}\left(\frac{1}{2}-\frac{1}{1+a^n}\right) \qed. \end{aligned}
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