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

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We need to find the value of lim n--> inf [ bn/2^n]

Also, we have bn = 1 + 2 + 2^n

lim n--> inf [ bn/2^n]

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

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

for n--> inf , we have 2^n --> inf and (1/2^n) --> 0

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

=> 0 + 0 + 1

**The required limit = 1**

[This answer is the solution to the unedited question.]

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.**