Simplifying a sum over nested summations and products of prime divisors

[drewfstr314]

Is there any way to simplify

$\sum_{n=1}^{\eta} \left(\sum_{p|n} \frac{1}{p} + \frac{1}{\prod_{p|n}p} \right)$ for a known η, and where "p|n" is a prime that divides n, i.e. p is a factor of n?

Thanks

 

[chiro]

Hey drewfstr314 and welcome to the forums.

For your p|n terms, do you get all prime factors or only the one of the prime factors?

For example with a number N = 2^2*3*5*7^2 all factors include the set {2,2,3,5,7,7} but if we only consider the primes themselves we get {2,3,5,7}.

If we had the first case we can simplify a great deal by taking the right-most term to be 1/n but otherwise we might have to resort to a special analytic function.

With regard to the left-most term again if we assume the above, we can collect these terms and get an expression in terms of X/Y where Y = N in the same way we do 1/a + 1/b = (b+a)/ab.

Now of course you do have some special functions in analytic number theory that could be used, but they tend to be complicated and I'm not sure what your goal is for using such a function.

Maybe you could elaborate on what you are trying to do so that the readers can give more specific advice.

 

[haruspex]

Is there any way to simplify

$\sum_{n=1}^{\eta} \left(\sum_{p|n} \frac{1}{p} + \frac{1}{\prod_{p|n}p} \right)$ for a known η, and where "p|n" is a prime that divides n, i.e. p is a factor of n?

For the first term you can swap the order of summation:

$\sum_{n=1}^{\eta} \sum_{p|n} \frac{1}{p} = \sum_{p\le\eta} \sum_{k=1}^{[\eta/p]} \frac{1}{p} = \sum_{p\le\eta} \frac{[\eta/p]}{p}$
No ideas for the second term.