I can't understand how the infinity concept plays into this.

Question

The value of the infinite product: 2^1/5 * 4^1/25 * 8^1/125 * 16^1/625 * ... can be expressed in simplest radical form as the n^th root of a to the b power. Compute n+a+b.

 

I got up to the point where I know that the factors in the equation go in the formula 2^n/5n, and I converted it to radical form, but I still don't know what n, a, and b are. Thanks!

Answer 1

You are dealing with an infinite product.    Call the infinite product Q.    Convert the infinite product to an infinite sum by taking the log₂ of both sides.     This results in:

 

log₂(Q) =   ∑ _n=1^∞  n/5ⁿ  

 

This infinite sum can be found on Wikipedia.     The infinite sum is 1/(1-1/5)² =  25/16.   So

 

log₂(Q) =   25/16          Both side can be exponentiated with 2  to get

 

Q =   2^25/16     This is the 16th root of  a^b   where a =2 and b = 25

 

So finally n =16,  a = 2 and b = 25.     Thus n+a+b =  43.

 

 

Answer 2

the product is infinite so we have 2^{a + b + c + d.....}

where a = 1/5, b = 2/25, c = 3/125, etc

 

in general the terms are n/5ⁿ

 

So the power of 2 is ∑ n/5ⁿ as n goes from 1 to ∞

 

This infinite series converges. The question is what is the value of this infinite sum.

I don't know if you are asking as a student in algebra 2 or calculus class.

 

If the sum = b/n and a = 2, then the infinite product = n^th root of 2^b

 

Perhaps a graphing calculator can give you the value for the infinite sum.