# Series

• June 5th 2011, 11:24 PM
Blizzardy
Series

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)
• June 6th 2011, 02:45 AM
CaptainBlack
Quote:

Originally Posted by Blizzardy

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)

\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}