In the infinite sequence
a, b, (a + b)/2, (a + 3b)/4, . . .
each term after the second is the arithmetic mean of the two previous terms.
Find the limit of the sequence in terms of real numbers a and b.
Let us denote the nth term of the give sequence as S(n).
Then, we observe that:
S(1) = a
S(2) = b
S(3) = a/2 + b/2
S(4) = a/4 + 3*(b/4)
S(5) = 3*(a/8) + 5*(b/8)
S(6) = 5*(a/16) + 11*(b/16)
S(7) = 11*(a /32) + 21*(b/32)
S(8) = 21*(a/64) + 43*(b/64)
S(9) = 43*(a/128) + 85*(b/128), and so on.
It can be easily shown by means of analytical procedure that;
S(2j) = [1- (1/4)^(j-1)]*(a/3) +[2 + (1/4)^(j-1)]*(b/3); and:
S(2j - 1) = [1+ (1/2)^(2j-3)]*(a/3) +[2 - (1/2)^(2j-3)]*(b/3)
whenever j = 1,2,3, 4, ....
In other words,
S(p)
= [1- (1/4)^(p/2 -1)]*(a/3) +[2 + (1/4)^(p/2-1)]*(b/3);
whenever p is even
= [1+ (1/2)^(p-2)]*(a/3) +[2 - (1/2)^(p-2)]*(b/3);
whenever p is odd
Accordingly,
Limit (p-> infinity) S(p) = a/3 + 2*(b/3)
Consequently, the required limit is (a+2b)/3
Solution
