Answer to Question #2330 in Real Analysis for alveen

Question

1. Consider the sequence
x1 = 1, x2 = 1 - 1/2, x3 = 1 - 1/2 + 1/4, x4 = 1 - 1/2 + 1/4 - 1/8...

(a) Show that:

(i) The subsequence x1; x3; x5; : : : is decreasing.

(ii) The subsequence x2; x4; x6; : : : is increasing.

(iii) Each even-numbered term is less than each odd-numbered term.

Answer 1

(i) Notice that the subsequence
x1; x3; x5….
Can be given by the following formula:
has the following consists of sums of geometric progression with first element 1/2 and
quotient 1/4:
x1=1
x3 =1-(1/2-1/4) = 1-1/4
x5= 1-(1/2-1/4)-(1/8-1/16) = 1-1/4-1/16
x7= 1-1/4-1/16 - 1/64 =
1-1/4(1+1/4+1/4^2)
…
x_{2n+1} =1-1/4(1+1/4+1/4^2+…+1/4^{n-1})
This shows that the sequence decreases
(ii) Notice that the subsequence
x2; x4; x6….
consists of sums of geometric progression with first element 1/2 and quotient 1/4:
x2=1-1/2=1/2
x4 =(1-1/2)+(1/4-1/8)=1/2+1/8
x6= 1/2+1/8+1/32…
…
x_{2n} =
1/2(1+1/4+1/4^2 +…+1/4^n)
This shows that the sequence increases.
(iii) Notice that the sum of geometric progression
1+1/4+1/4^2+… = 1/(1-1/4) = 4/3
Hence every even term is less than
1/2 * 4/3 = 2/3
while every odd term is greater than
1-1/4 *4/3 = 2/3
Thus
x_2n < 2/3 < x_2k+1
for any n,k >0