dn = 1 + 2 + 4 + 8 ...2^n

dn / 2^n = 1/2^n + 1/2^n-1 +...1

Now lim n--> inf. [ dn / 2^n]

=> lim n-->inf. [ 1/2^n + 1/2^n-1 +...1]

We know that lim n-->inf. [ 1/2^n ] = 0 and as n-->inf., all the terms in the above series tend to zero except 2^n/2^n which is equal to 1.

=> lim n-->inf. [ 1/2^n + 1/2^n-1 +...1]

=> 0 + 0 + 0 ...+ 1

=> 1

**The required result is 1.**

We notice that the terms of the sum are the terms of a geometric progression, whose common ratio is r = 2.

The sum of n terms of a geometric progression is:

dn = 1*(2^n - 1)/(2 - 1)

dn = (2^n - 1)

Now, we can evaluate the limit;

lim dn/2^n = lim (2^n - 1)/2^n

We'll get:

lim (2^n)/2^n - lim 1/2^n

We'll simplify and we'll get:

lim 1 - lim 1/2^n

since n->+infinite => lim 2^n = infinite => lim 1/2^n = 0

**lim dn/2^n = 1 - 0**

**lim dn/2^n = 1, for n -> +infinite**