# Answer to Question #2330 in Real Analysis for alveen

Question #2330

1. Consider the sequence

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

<br>(a) Show that:

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

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

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

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

<br>(a) Show that:

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

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

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

Expert's answer

(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

for any n,k >0

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

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