1. Consider the sequence: a1 =?6 a2 =sqrt (6+sqrt6) a3 =sqrt(6+sqrt6+sqrt6) a4 =
ID: 2828630 • Letter: 1
Question
1. Consider the sequence:
a1 =?6
a2 =sqrt (6+sqrt6)
a3 =sqrt(6+sqrt6+sqrt6)
a4 =sqrt(6+sqrt6+sqrt6+sqrt6)
. . .
(a) Find a recursion formula for an+1.
(b) Use mathematical induction to prove that an < 4 for all n = 1,2,3,....
(c) Use mathematical induction to prove that the sequence is increasing.
(d) Because you have proved that the sequence is bounded above and increasing, it must have a limit. find the limit.
for part a) i got
a_(n+1)= sqrt(6+a_n-1)
and thats it, im stuck, please help! i'll be generous with points.
Explanation / Answer
a) Using your notation, a_(n+1) = sqrt(6+a_n)
A key fact used in b and c is that the square root is an increasing function.
b) For n = 1 sqrt(6) < 4
Assume a_k < 4
Then, a_(k+1) = sqrt(6 + a_k) < sqrt(6 + 4) = sqrt(10) < 4
Then, a_n < 4 for all n.
c) For n = 1, a_1 = sqrt(6) and a_2 = sqrt(6 + sqrt(6)) > sqrt(6) = a_1, so a_2 > a_1
Assume for n = k. a_k+1 > a_k
Then, as a_k+1 > a_k, a_k+2 = sqrt(6 + a_k+1) > sqrt(6 + a_k) = a_k+1, so a_k+2 > a_k+1
d) To find the limit, let x = sqrt(6 + x)
Expanding both sides, x^2 = 6 + x
x^2 - 6 - x = 0
(x - 3)(x+2) = 0
x = -2, 3
Clearly, x is not negative, as a_n >= sqrt(6) for all n
Thus, x = 3
(If you want more proof, first change b) to show that a_n < 3 for all n: sqrt(6) < 3; assum a_k < 3; then, sqrt(6 + a_k) < sqrt(6 + 3) = sqrt(9) = 3
Then, lastly, if you assume any limit L less than 3, then there exists an n such that a_n > L - epsilon for any epsilon. Then, for any L < 3, you can find a sufficiently small epsilon so that sqrt(6 + L - epsilon) > L
The proof follows. square both sides: to show 6 + L - epsilon > L^2
as L < 3. say 3 - delta, and delta < 1 (as 3 - sqrt(6) < 1), so delta^2 < delta
L^2 = 9 - 6 delta + delta^2 < 9 - 5 delta, so 6 + L - epsilon = 6 + 3 - delta - epsilon = 9 - delta - epsilon > 9 - 5 delta (as long as epsilon < 4 delta) > L^2. This contradicts L being the limit, so L must equal 3.
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