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

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 nth 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 2n/5n, and I converted it to radical form, but I still don't know what n, a, and b are. Thanks!
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2 Answers

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 =   225/16     This is the 16th root of  ab   where a =2 and b = 25
So finally n =16,  a = 2 and b = 25.     Thus n+a+b =  43.
the product is infinite so we have 2a + 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 = nth root of 2b
Perhaps a graphing calculator can give you the value for the infinite sum.