| Here is my solution, proving b) first, then a).
1. For each n, u_n is well defined and 0 < u_n < 1.
This is true for n = 0 and n = 1 by definition. And by induction
we have for n >= 2:
u_n = (sqrt(u_(n-1)) + sqrt(u_(n-2))/2 > 0
and
u_n = (sqrt(u_(n-1)) + sqrt(u_(n-2))/2 < (sqrt(1) + sqrt(1))/2 = 1
2. The sequence u_n is (strictly) increasing.
Let's compute for an n >= 2:
u_n - u_(n-1) = - u_(n-1) + 1/2 sqrt(u_(n-1)) + 1/2 sqrt(u_(n-2))
In order to know the sign of this expression, we have to look at
the following function (for a fixed n and a variable x):
x |--> f(x) = -x� + 1/2 x + 1/2 sqrt(u_(n-2))
The roots of the equation f(x) = 0 are:
x_1 = ( 1 - sqrt( 1 + 8 * sqrt(u_(n-2) ) ) / 4
x_2 = ( 1 + sqrt( 1 + 8 * sqrt(u_(n-2) ) ) / 4
For each x in the interval [x_1, x_2], we have f(x) > 0 (the opposite
sign of the leading coefficient) and f(x) < 0 for each x outside this interval.
Since f(0) = 1/2 sqrt(u_(n-2)) > 0 and
f(1) = 1/2(1+sqrt(u_(n-2)) > 0, we conclude that f(x) > 0 for each
x in [0, 1]. (In other words, the interval [0, 1] is included in
the interval [x_1, x_2]).
Since u_(n-1) is in [0, 1], also sqrt(u_(n-1)) is in [0, 1] and
u_n - u_(n-1) = f( sqrt ( u_(n-1) ) ) > 0.
3. The real sequence u_n has an upper bound ( =1 ) and is increasing.
Then it has a limit l. For continuity reasons (of the square root function),
this limit must verify:
l = 1/2(sqrt(l) + sqrt(l)), or l = sqrt(l), so l = 0 or l = 1.
Since u_n >= u_2 > 0 for each n >= 2, l > 0, so l = 1.
I don't see a simple way to prove a) first then b).
Herv�
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| >2. The sequence u_n is (strictly) increasing.
I'm suprised there's no simpler way of proving this (and I've looked).
>3. The real sequence u_n has an upper bound ( =1 ) and is increasing.
>Then it has a limit l.
This isn't clear. For example, the sequence:
v_1 = 0, v_2 = 1/4, v_n = v_<n-1> + 2^(-n)
has an upper bound of 1 and is increasing. But that doesn't make its limit 1.
(FWIW, the v sequence's _least_ upper bound is 1/2, as is its limit).
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| Let's make it clear: I said (and it's true) that a sequence which has
an upper bound (in this case the real number 1, one) and is increasing
has a limit (let's call it l, the letter 'l' for 'Linda', 'Louise' or
whatever you like).
Since the bound is 1 (one) all we can say is that l
(letter) is less or equal to 1. But the limit of u_(n-1) is l, also the
limit of sqrt(u_(n-1)) is sqrt(l), and same for u_(n-2). So the limit
of 1/2(sqrt(u_(n-1))+sqrt(u_(n-2)) is 1/2(sqrt(l)+sqrt(l))=sqrt(l).
But 1/2(sqrt(u_(n-1)+sqrt(u_(n-2)) is u_n by definition, whose limit is
l (for 'Laurence': it's a female name in French), so l = sqrt(l) and
then l is 0 or 1. It cannot be 0 since all the terms of the sequence
are greater than the term u_2 which is *strictly* greater than 0.
Herv�
BTW, I feel a little ashamed of the wrong proof in .1: I wasted my
week-end :-)
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