Sequence of real numbers

kingwinner

我 prove 左 part b, 但唔識點 prove parts a & c...

請高手賜教, thx!

Naozumi

(a) Suppose 0< an < 1 for all n
     a(n+2) > sqrt(a(n+1))
Since 0< an < 1, sqrt(a(n+1)) > a(n+1)
Hence,  a(n+2) > a(n+1)
So {an} is an increasing sequence of positive numbers with upper bound 1
i.e. lim an exist, say, a
so a = sqrt(a) + sqrt(a)
     a = 2sqrt(a)
    a = 0 or a = 4
Recall that {an} is an increasing sequence of positive numbers with upper bound 1, hence, neither 0 nor 4 can be the limit of the sequence.

[ 本帖最後由 Naozumi 於 10-1-27 11:32 PM 編輯 ]