Divider Natural
help me with a sum..or suggest something?
we define f(n) as the number of different ways that we can write natural number n like a product of two co-prime numbers..example 6:=3.2=6.1 so f(6)=2, or f(4)=1,(3.2 and 2.3 is like one way)
now we define r(n) as the number of different prime dividers of natural n..example: 6/3=2,6/2=3 so r(6)=2..or r(7)=1
find:
∞
Σ[(-1)^r(n)].[f(n)^3]/(n^3)
n=1
be careful that exercise is hard..so don't waste your time if you are not experienced with difficult exercises..thanks
of course and there is a way to find it since that is an exercise that i have to solve..but if you can at least to show that it converges just write it down,if nobody will answer to me i will choose it as best answer..
i'm sorry..the division it has to be /(n^2) and not /(n^3)..
Edit: The asker has solved it himself (good for you!) and has spotted a mistake in my work and finished the last part of the argument. So I will just amend my work here for others' benefit.
First let's relate f(n) to r(n). Consider the prime factorization of n into prime powers. There are clearly r(n) such prime powers by definition of r(n) as the number of prime divisors. Now notice that each time we write n as a product of coprime numbers a and b, a and b cannot share prime divisors. This means of course that a must be the product of some prime powers of n and b must be the product of the remaining prime powers. a or b can of course also be 1, in which case the other one has all the prime powers. So to count the number of ways to express n as ab we just have count the number of ways to split the prime divisors of n into two subsets, which is nothing but 2^r(n). But we do not care about the order of a and b, so f(n) = [2^r(n)]/2.
So now, the required sum is
Σ[(-1)^r(n)].[f(n)^3]/(n^2)
= Σ[(-1)^r(n)].[[2^r(n)]/2]^3/(n^2)
= (1/8)Σ[(-1)^r(n)].[2^r(n)]^3/(n^2)
= (1/8)Σ[(-8)^r(n)]/(n^2)
Now observe that the infinite sum can be expressed as an infinite product over all primes p:
(7/8) + (1/8)Π(1 + (-8)/p^2 + (-8)/p^4 + ...)
Here's the reason why this product works out to be the original sum: when you expand out the product, each term in the original sum, say [(-8)^r(n)]/(n^2), is reconstituted from r(n) factors of the form (-8)/p^(2i).
So, continuing with the product, we simplify it as follows:
(7/8) + (1/8)Π(1 + (-8)/p^2 + (-8)/p^4 + ...)
= (7/8) + (1/8)Π(1 + (-8)[1/p^2 + 1/p^4 + ...])
= (7/8) + (1/8)Π(1 + (-8)[-1 + 1 + 1/p^2 + 1/p^4 + ...])
= (7/8) + (1/8)Π(1 + (-8)[-1 + 1/(1-1/p^2)])
= (7/8) + (1/8)Π(1 + (-8)[-1 + p^2/(p^2 - 1)])
= (7/8) + (1/8)Π(1 + (-8)[1/(p^2 - 1)])
= (7/8) + (1/8)Π(1 - 8/(p^2 - 1))
I must admit I'm stuck at this point. But you may know how to use this info or continue the argument.
And this is what the asker told me that would finish the argument: at p = 3, 1 - 8/(p^2 - 1) = 0 and so the whole infinite product vanishes, leaving 7/8. Cool!
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