Math: Sequence need help please?

Question

1. Define the sequence {Sn } inductively by: S1 = square root of( 3) and Sn+1 = square root of (3 Sn).
(a) Prove Sn is bounded above by 3.
(b) Prove Sn is increasing
(c)Deduce that Sn converges


2. Let S be a set of real numbers such that S is not equal empty set and S is bounded below. Thus
(There exists u in R) ( x in S => u Let T = { x: -x in S}
(a)Prove that T is not equal to empty set and that T is bounded above by –u
(b) Let b = l.u.b.T, Prove –b = g.l.b.S

Answer 1

i'll help you get started:

it's pretty easy to "see" that S1=3^(1/2), then S2=3^(3/4), then S3=S^(7/8)..

more generally, Sn = 3^((2^n-1)/2^n)

as n grows large, the exponent tends towards 1 from below. As the function 3^x is strictly growing (positive derivative) for positive integers, Sn tends towards 3^1 = 3 and is bounged above by it

for (b) you can either compute the difference between two consecutive terms from the Sn(n) I gave you above. it will always be positive.

...
good luck

Answer 2

1) So S1 = 3^(1/2), and S(n+1) = (3*Sn)^(1/2) = 3^(1/2)Sn^(1/2).

The first few terms would be S1 = 3^(1/2), S2 = 3^(1/2)3^(1/4), S3 = 3^(1/2)3^(1/4)3^(1/8), etc. This can be rewritten as Sn = 3^[(1/2)^1 + (1/2)^2 + (1/2)^3 + ... + (1/2)^n].

a) The term in the exponent of Sn (as rewritten above) is increasing but has a limit of 1. Therefore, Sn can't be greater than 3^1, or 3.
b) The terms of the exponent of Sn keep increasing with each n, therefore so does 3 to that power. Thus Sn increases
c) The exponent converges to 1, so Sn converges to 3.

2) I might be missing something from the definition here, but it would seem that S = {1, 2, 3, 4, 5...} would be a set that fits the definition, since the numbers are real and it's bounded below. But then T would actually be the null set, since there is no number x in S such that -x is in S too. You sure it's not "bounded above"? Is

Answer 3

S[1] = sqrt(3).
S[n+1] = sqrt (3 S[n])

2) To prove that S[n] is increasing, we use induction. We want to prove that S[n+1] > S[n] for all n >= 1.

1) Let n = 1.
Then S[1 + 1] = S[2]
= sqrt( 3 S[1] )
= sqrt (3 sqrt(3) ), and
S[1] = sqrt(3).

We can verify with a calculator that S[2] > S[1].

2) Assume the formula holds true for n = k. That is

S[k + 1] > S[k].

{We want to prove that S[k + 2] > S[k + 1]}

Multiplying both sides by 3, we get

3 S[k + 1] > 3 S[k]

Taking the square root of both sides,

sqrt (3 S[k + 1]) > sqrt (3 S[k]), which means

S[k + 2] > S[k + 1]

Therefore, the formula holds true for n = k + 1, so by the principal of mathematical induction, S[n + 1] > S[n].