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# A sequence is defined recursively by a1 = 1 and for n ≥ 1, an+1 = √(3an). Given that this sequence converges two?

A sequence is defined recursively by a1 = 1 and for n ≥ 1, an+1 = √(3an). Given that this sequence converges to a positive number, find the limit of the sequence.

(a) √3

(b) 1/3

(c) 3

(d) 6

(e) 2/3

Can someone please explain how to do this? Thanks

### 2 Answers

- RealProLv 71 month agoFavorite Answer
a2 = √3 * √(a1)

a3 = √3 * √(a2)

= √3 * √(√3 * √(a1))

= √3 * √(√3) * √(√a1)

a4 = √3 * √(a3)

= √3 * √(√3 * √(√3) * √√a1)

= √3 * √(√3) * √(√(√3)) * √(√(√a1))

= 3^(1/2) * 3^(1/4) * 3^(1/8) * (a1)^(1/8)

= 3^(1/2 + 1/4 + 1/8) * (a1)^(1/8)

Clearly the subsequent terms will have even more nested square roots, adding powers of 2.

a_[n+1] = 3^(1/2 + 1/4 + 1/8 + 1/16 + .... + 1/2^n) * (a1)^(1/2^n)

As n goes to infinity, 1/2 + 1/4 + 1/8 + 1/16 + .... + 1/2^n tends to 1 while (a1)^(1/2^n) tends to 1 (the infinity-th root of a1)

3^(1) * 1 = 3

This is a proof that it converges as well as the value of the limit.

If you already know it converges like you do, then just use the definition of a limit.

As sufficiently large n, a_n and a_[n+1] are arbitrarily close. So close, in fact, that we can consider them the same number.

a_n = a_[n+1]

a_n = √(3a_n)

L = √(3L)

L^2 = 3L

L = 0 or L = 3

L = 0 is the solution for the case a1 = 0. For all other positive a1, the limit is 3.

- Φ² = Φ+1Lv 71 month ago
a_1 = 1 = 3^(0/1)

a_2 = 3^(1/2)

a_3 = 3^(3/4)

a_4 = 3^(7/8)

etc.

Limit = 3^((2^n - 1)/2^n) = 3 as positive n becomes large.

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Thanks for the great explanation!