Pure Maths --- Sequences

其實不需用到 MI 的, 看以下的 deduction:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Mar08/Crazystevig8.jpg

當然, 這仍需用到 a, b are positive 的 assumption.
From the result above, we can deduce that the sequences:
{x1, x3, x5, ...} and {x2, x4, x6, ...} are both monotonic decreasing.

根據你條問題，A和B應該係 positive integers。
而且，你要證明 x[1]細過x[3]細過x[5]....... 及 x[2] 細過 x[4] 細過 x[6]........
你係咪打錯題目？

我打唔到[大過]和[細過]符號，所以用中文代替兩個符號。

from the given function,
x[1]=b
x[2]=a/b,
x[3]=b/2 [細過] x[1]
x[4]=2a/3b [細過] x[2]

So, x[1] [細過] x[3] & x[2] [細過] x[4]

x[k 1]
= a/kx[k]

x[k 2]
=a/(k 1)x[k 1]
=kx[k]/(k 1) [細過] x[k] when k [大過] 0

So, x[k 2] [細過] x[k] for all ve intergers k

By M.I , it is true for all ve integers k.

簡單黎講，第三個大過第一個，第四個大過第二個。而對所有K，第K 2個大過第K個。所以一次過 PROVE 兩個都 decreasing

第二個唔知是否大過第一個，因為唔知a,b係咩。