What's known about $\sum_{n=1}^\infty\frac{1}{\sigma_s(n)}$?

Question

Let $$\sigma_s(n)=\sum_{d|n} d^s$$

$$f(s)=\sum_{n=1}^\infty\frac{1}{\sigma_s(n)}$$

(1) Is it possible to prove that $f$ converges for $s>1$?

(2) Is there anything that can be said about an analytic continuation of $f$? Namely, is there a unique analytic continuation to $\mathbb{C} \setminus\{1\}$?

Here's what's clear to me: $f$ diverges at $s=1$. Let $\mathbb{P}$ denote the prime numbers.

$$f(s)> \sum_{p\in\mathbb{P} } \frac{1}{\sigma_s(p)}=\sum_{p\in\mathbb{P}}\frac{1}{1+p^s}$$ And this last expression converges iff $P(s)$ the prime zeta function converges. So in particular $f(s)$ is divergent at $s=1$.

Answer 1

For (1), the answer is yes, since $\sigma_s(n) \geq n^s$, so $$\sum_n 1/ \sigma_s(n) \leq \sum_n n^{-s} < \infty$$

Answer 2

There is of course no meromorphic continuation to the whole complex plane. Expanding $$1+\sum_{k\ge 1}\frac{p^{-sk}}{1+\sum_{m=1}^k p^{-ms}}=1+\sum_{l\ge 1} a(l) p^{-sl}=f(p^{-s})$$ We find that your function is the Dirichlet series with multiplicative coefficients $$F(s)=\sum_n \sigma_s(n)^{-1}=\prod_p (1+\sum_{l\ge 1} a(l) p^{-sl})=\sum_n b(n)n^{-s}$$ Let $$\log f(z)= \sum_l c(l) z^l$$ which is analytic for $|z|<1$ thus $$\log F(s)-\sum_{ld< L}c(l)\frac{\mu(d)}{d} \log \zeta(lsd)$$ is analytic for $\Re(s) > 1/L$ which means that $F(s)^L$ has a meromorphic continuation to $\Re(s) > 1/L$ with some zeros/poles at $1/l$ and $\rho /l$.

$F(s)$ has a natural boundary on $\Re(s)=0$.

The $\sum_{ld= L}c(l)\frac{\mu(d)}{d}=u_L/v_L$ are not integers thus $F(s)$ has some $v_L$-rooth branch points at $1/L$ and $\rho/L$.