**Any power of 2 greater than 8 will work.**

For example `A(16)=(1+2+4+8+16)/16=31/16=1.9375`

** The powers of two form a geometric progression; each term is the product of the previous term and 2. All of the factors of `2^n` are the powers of 2 from 0 to n. e.g. the factors...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

**Any power of 2 greater than 8 will work.**

For example `A(16)=(1+2+4+8+16)/16=31/16=1.9375`

** The powers of two form a geometric progression; each term is the product of the previous term and 2. All of the factors of `2^n` are the powers of 2 from 0 to n. e.g. the factors of 128=`2^7` are `1=2^0,2=2^1,4=2^2,8=2^3,16=2^4,32=2^5,64=2^6,128=2^7`

An alternative description for the abundancy ratio is the sum of the reciprocals of the factors. Thus `A(16)=1/1+1/2+1/4+1/8+1/16=1.9375=(1+2+4+8+16)/16`

The sum of a finite number of terms of a geometric progression is given by `a_1(1-r^n)/(1-r)` where `a_1` is the first term and `r` is the common ratio and `n` the number of terms.

Since for this problem we are considering powers of two, the reciprocals of the factors form a geometric progression with `a_1=1,r=1/2` . The the finite sum for any n is `(1-(1/2)^n)/((1/1/2))=2(1-(1/2)^n)` It is clear that this is less than 2 for any n, and `2(1-(1/2)^n)>1.9==>1-(1/2)^n>.95==>(1/2)^n<.05==>n>3`

Thus any power of two greater than 8 will work.

**Further Reading**