# Given the sum : 1 + 1/2 + 1/2^2 + ... + 1/2^100, prove that 2 >S >1

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### 4 Answers

S = 1 + 1/2 + 1/2^2 + ...+ 1/2^100

We notice that 1, 1/2, 1/4, 1/8 , ...., 1/2^100 are parts of a geometric progression swhere r = 1/2

==> We know that:

Sn =( r^n)-1)/ (r-1) = (1-r^n)/(1-r)

S101 = (1/2)^101 -1]/(-1/2)

= -2(1/2)^101 -1]

= 2- (1/2)^100

But we know that:

1>(1/2^100) >0

Now multiply by -1

==> -1<-(1/2^100) < 0

Now add 2:

==>1 <2- 1/2^100 < 2

==> 1<Sn< 2

The given sum is the sum of the terms of a geometrical progression. The number of terms is 101.

We'll calculate the common ratio of the progression:

1/2/ 1 = 1/2

1/4/1/2 = 1/2

.....................

r= 1/2.

We know that the sum of n terms of a geometrical progression is:

Sn=(r^n-1)/(r-1), when r>1, and Sn= (1-r^n)/(1-r), r<1

We've noticed that r=1/2<1 and n=101

S101 = (1-1/2^101)/1-1/2=2 - 1/2^100

1/2^100 < 2, so S101>1 and S101<2.

The series given is a GP with the first term as 1 and the common ratio as 1/2. For a GP with common ratio less than 1, the sum of the first n terms is (1-r^n)/(1-r)

Now we need to find the sum of 101 terms as 1/2^100=1/2^(101-1) and for a GP the nth term is ar^(n-1)

The sum= (1-r^n)/(1-r)=(1-(1/2)^101)/(1/2)= (2)-(1/2)^100

Now as (1/2)^100 is greater than 0, the sum is less than 2.

Also, as (1/2)^100 is less than 1 the sum is greater than 1.

Let the sum S = 1+1/2+1/2^2+....1/2^100, the sum of 101 termr of the GP.

S = 1+x+x^2....+x^n, where 1 is 1st term = 1/2, the common ratio, and xn is the n+1th term of the

= (1-x^n+1)/(1-x)

= 1/(1-x) + x^n/(1-x)

= 1/(1-1/2) - x^n+1 / (1-1/2)

= 2 - (1/2)^(100+1)/2 <2.

So 1 < S < 2.