Thread: 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)

Originally Posted by Blizzardy

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

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.