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

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.