What is the limit of the fraction bn/2^n if bn=1+2+2^n, n-->infinite?

In other words, we'll have to evaluate the limit of the
ratio:

lim (1 + 2 + ... + 2^n)/2^n,
n->infinite

We'll apply the Cesaro-Stolz's theorem
and we'll determine:

bn+1 - bn = [1 + 2 + ...+ 2^n +
2^(n+1)]- (1 + 2 + ... + 2^n)

We'll eliminate like terms
and we'll get:

bn+1 - bn =
2^(n+1)

The limit of the function bn/2^n = lim (bn+1 -
bn)/(2^(n+1) - 2^n)

lim (bn+1 - bn)/(2^(n+1) - 2^n) = lim
2^(n+1)/(2^(n+1) - 2^n)

We'll factorize the denominator by
2^(n+1):

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

We'll simplify and we'll
get:

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

lim (1 + 2 + ... + 2^n)/2^n =
2

The limit of bn/2^n =
2.