# Summing the Riemann zeta function over divisors of a natural number

In one of my computations the following term appeared for $n\in\mathbb{N}$: $$f(n) := d(n) - \sum_{k\geq 2, k\mid n}\zeta(k)$$

Here $d$ denotes the number-of-divisors function: $d(n)=\sum_{k\in\mathbb{N}, k\mid n}1$ and $k\mid n$ means $k$ is a divisor of $n$ as usual.

I am working in topology, so my experience in number theory is rather limited. Despite a thorough search in literature and online, I could not find any reference to such a sum over values of the Riemann zeta function appearing in number theory, let alone the combination with the number-of-divisors function. Can anybody provide me with a reference to a known appearance of such a sum in number theory if it exists? What could be possible applications of such a sum in number theory?

Here is what I can show easily about the expression $f(n)$ from my computations:

1) It is bounded by $$0<f(n)<1, \forall n\in\mathbb{N}$$
where both bounds a sharp.

2) It is sub-additive: $$f(n+m)\leq f(n)+f(m), \forall m,n\in\mathbb{N}$$

Are these results known about such a combination of the Riemann zeta function and the number-of-divisors function in number theory? If yes, are they obvious?

• How do you get this weird sum $\sum_{d | n,d \ge 2} \zeta(d)$ ? And use another symbol, since $\sigma(n)$ means $\sum_{d | n} d$ – reuns Nov 20 '16 at 0:58
• True. I changed $\sigma$ to $f$. – AlexD Nov 20 '16 at 10:59

• All you need is that $|\zeta(x)-1| < C 2^{-x}$ for $x \ge 2$, so with $f(n) = d(n)- \sum_{k |n,k\ge 2} \zeta(k)$ you get $$0 \le 1-f(n) = \sum_{k |n,k\ge 2} (\zeta(k)-1) < \sum_{k=2}^\infty C 2^{-k} = \frac{C2^{-2}}{1-2^{-1}} = C/2$$ Now for $x \ge 2$, since $n^{-x} < \int_{n-1}^n t^{-x}dt$ : $$0 <\zeta(x)-1 =2^{-x}+\sum_{n=3}^\infty n^{-x}< 2^{-x}+ \int_2^\infty t^{-x}dt = 2^{-x} + \frac{2^{-x}}{x-1} \le 2^{1-x}$$ i.e. $C = 2$ works and you get $1-f(n) \in [0,1]$ and $f(n) \in [0,1]$.