On the complex function $f(s)=\sum\limits_{n=1}^\infty\sigma(n)^{-s}$ Let $s=x+iy$ the complex variable (if you want a difffernt notation you are welcome), then I know that $\sum_{n=1}^\infty n^{-s}$ converges for $\Re s>1$. On the other hand let $\sigma(n)=\sum_{d\mid n}d$ the sum of divisors function. An important fact is that $\sigma(p)=p+1$ if and only i $p$ is a prime number. We define for some abscissa of convergence $$\sum_{n=1}^\infty\frac{1}{(\sigma(n))^s}.$$
Since applying the triangle inequality one has $$\left|\sum_{n=1}^\infty\frac{1}{(\sigma(n))^s}\right|\leq\sum_{n=1}^\infty\frac{1}{|\sigma(n)^s|}=\sum_{n=1}^\infty\frac{1}{\sigma(n)^{\Re s}},$$
and one knows that $\sigma(n)\geq n+1>n$ for each $n>1$ (notice that $\sigma(1)=1$) then $(\sigma(n))^{-\Re s}<n^{-\Re s}$ with convergence of the infinite series thus for $\Re s>1$ by the comparison test. And since $\sum_{p \text{ prime}}1/p$ diverges (thus by comparison also $\sum_{p \text{ prime}}1/(p+1)$) I believe that the abscissa of convergence is $1$ (please if there is some mistake or inaccurancie in my claims say me, I want learn and write mathematics rigurously). 
Thus the information that I can deduce for this function is that has a pole at $s=1$ and converges in the half-plane $\Re s>1$. But I don't know what different claims can be deduced easily.

Question. I would like to learn more about complex analysis, what's about this function $$f(s)=\sum_{n=1}^\infty\frac{1}{(\sigma(n))^s}$$
  concerning if  is it meromorphic, (I hope that my words and claims are rights), it is possible/feasible an analytic continuation to different regions of the complex plane, or can you deduce other easy things about non-vanishing for $\Re s>1$...? Has zeros? 

Thus I am asking about what's things are easily deduced for a complex function of this kind. I understand that is a question involving a lot of possible computations, but I would like to know how works with this kind of functions from the complex path. You can deduce the more relevant facts and other provide us as hints.
Thanks in advance.
 A: This function extends meromorphically to $\Re(s)>0$ (with a single pole of order $1$ at $s=1$), and has at least the same zeroes as the Riemann zeta function does there.
Indeed, by using multiplicativity it follows that for $\Re(s)>1$ we have the following formula:
$$f(s) = \prod_{p} (1+\frac{1}{\sigma(p)^s}+\frac{1}{\sigma(p^2)^s}+...)$$
This is already enough to see that this function has no zeroes for $\Re(s)>1$: by using $\sigma(p^k) > p^k$, one shows this each factor is $1+O(\frac{1}{p^s})$, and each factor is non-zero. Such fast-converging products are known go arrive at a non-zero value.
Let me first show the how to extend it for $\Re(s)>1/2$, and the general extension will follow from a more careful analysis. Because $\sigma(p) = p+1$, and  $\frac{1}{(p+1)^s} = \frac{1}{p^s} + O(\frac{1}{p^{s+1}})$ each factor is equal to:
$$1 + \frac{1}{p^s} + O(\frac{1}{p^{2s}})$$
(notice that the contribution of the powers of $p$ is at most $O(1/p^{2s})$, which dominates $\frac{1}{p^{s+1}}$ for $s<1$).
Now divide this function by the zeta function and use the zeta function Euler product to get that for $\Re(s)>1$:
$$\frac{f(s)}{\zeta(s)} = \prod_p(1+\frac{1}{p^s}+O(p^{2s}))(1-\frac{1}{p^s}) = \prod_p(1+O(\frac{1}{p^{2s}}))$$
However this product is defined and is holomorphic (and non-zero) not only for $\Re(s)>1$, but for $\Re(s)>1/2$, so the ratio $\frac{f(s)}{\zeta(s)}$ is holomorphic there, so $f$ extends holomorphically to that region and has the same zeroes and poles as $\zeta$ there.
One can show in fact that the factors of the ratio between $f$ and $\zeta$ satisfy: 
$$(1-\frac{1}{p^s})(1+\frac{1}{\sigma(p)^s}+\frac{1}{\sigma(p^2)^s}+...) = 1+O(\frac{1}{p^{s+1}})$$
Which converges absolutely for $\Re(s)>0$, giving the desired extension.
I would be interested in hearing about further extensions of $f$ or reasons to expect there are none.
