sequence

We have to use the answer in a(ii)

(a_n+2 / a_n)
= (a_4 / a_2)^(2^(-(n-2))) when n is even
and (a_3 / a_1)^(2^(-(n-1))) when n is odd.

Consider n=2k. By c(i) a_2k converges to l.
Also, since n is even, we have
(a_n+2 / a_n) = (a_4 / a_2)^(2^(-(n-2))) = (a/b)^(2^-n)
(a_n / a_n-2) = (a/b)^(2^(-(n-2)))
(a_n-2 / a_n-4) = (a/b)^(2^(-(n-4)))
...
(a_4 / a_2) = (a/b)^(2^-2)

multiply all these together, and note that the terms in the R.H.S cancels out. We get

(a_n+2 / a_2) = (a/b)^(2^-2 + 2^-4 +...+ 2^-(n-2) + 2^-n))
Let n tends to infinity, L.H.S. tends to l/b, and
R.H.S. tends to (a/b)^((1/4)/(1- 1/4)) = (a/b)^(1/3)

Therefore,
l/b = (a/b)^(1/3)
l^3 / b^3 = a/b
l^3 = ab^2.

GOOD

重點是用了multiply all these together 和 a4/a2=(a/b)^1/4

我估到同a(ii)有關﹐但是想不到如何用﹐因為直接let n-> infinity只會出到1=1