Let $d(n)$ denote the divisor function. We know that, $d(n)= O(n^\epsilon)$ for any $\epsilon>0$. Let $k$ be a natural number. Is there any way to show $ \sum_{n=1}^{\infty} \frac{d(n)^k}{n^s}$ is absolutely convergent in $Res>1$ without assuming the fact that $d(n)= O(n^\epsilon)$ for any $\epsilon>0$? That is, can we prove this equality $ \sum_{n=1}^{\infty} \frac{d(n)^k}{n^s}= \prod_{p} (1+ \frac{2^k}{p^s}+ \frac{3^k}{p^s} + \cdots) $ converges for $Re s>1$?
I know that the function $\frac{d(n)^k}{n^s}$ is multiplicative. What are some other conditions necessary to imply the above equality holds.
I could only show the absolute convergence of the series with the assumption $d(n)= O(n^\epsilon)$ for any $\epsilon>0$. Any help or hint would be appreciated. Thanks in advance
Recall that if $a_n \geq 0$, then $$ 1 + \sum_{n = 1}^N a_n \leq \prod_{n = 1}^N (1 + a_n) \leq \exp\Big( \sum_{n = 1}^N a_n \Big),$$ and thus the convergence of the infinite product converges if and only if the infinite sum converges.
In this case, I note that as the coefficients $d(n)^k$ are real, the leading pole of the Dirichlet series $$ \sum_{n \geq 1} \frac{d(n)^k}{n^s} $$ will be real (this is sometimes called an old theorem of Landau). In fact, since $d(n) \geq 1$, we also know that the leading pole will satisfy $\mathrm{Re} s \geq 1$, but we don't use that here.
Thus to find the abscissa of convergence of the Dirichlet series, it suffices to consider only real valued $s$. For real valued $s$, the terms are all nonnegative. Note also that as $d(n)^k$ is trivially bounded by $n^k$ (assuming every number up to $n$ is a divisor of $n$), we find that the Euler product $$ \sum_{n \geq 1} \frac{d(n)^k}{n^s} = \prod_p \Big( 1 + \frac{2^k}{p^s} + \frac{3^k}{p^{2s}} + \cdots \Big) $$ agrees with the sum at least in the half-plane $\mathrm{Re} s > k + 1$, and both converge absolutely in that half-plane. To understand the half-plane of convergence further, we'll examine only the Euler product.
By the first note above, it suffices to consider convergence of the sum $$ \sum_p \Big( \frac{2^k}{p^s} + \frac{3^k}{p^{2s}} + \cdots \Big) = \sum_p \sum_{\ell \geq 1} \frac{(\ell + 1)^k}{p^{\ell s}}.$$ Again for $\mathrm{Re} s > k + 1$, this equality is true trivially and the RHS converges trivially. In this region, we can swap the order of summation to find that $$ \sum_p \sum_{\ell \geq 1} \frac{(\ell + 1)^k}{p^{\ell s}} = \sum_{\ell \geq 1} (\ell + 1)^k \sum_{p} \frac{1}{p^{\ell s}}. $$ We note the trivial bound (recalling that we restrict to real valued $s$, or alternately where we implicitly take absolute values around every summand in each expression) $$ \sum_{p} \frac{1}{p^{\ell s}} \leq \sum_{n \geq 2} \frac{1}{n^{\ell s}} \lesssim \int_2^\infty \frac{1}{t^{\ell s}} dt = \frac{2^{1 - \ell s}}{\ell s - 1}, \qquad (\mathrm{Re} s > 1)$$ where the integral converges if and only if $\ell s > 1$. As $\ell \geq 1$, the integral will converge for all $\ell$ if and only if $s > 1$. Thus $$ \sum_p \sum_{\ell \geq 1} \frac{(\ell + 1)^k}{p^{\ell s}} \lesssim \sum_{\ell \geq 1} \frac{(\ell + 1)^k}{2^{\ell s}}, \qquad (\mathrm{Re} s > 1) $$ and this last sum actually converges for all $s$ with $\mathrm{Re} s > 0$.
Tracing the sequence of inequalities, it follows that the Euler product representation converges absolutely for $\mathrm{Re} s > 1$, and thus the Dirichlet series converges absolutely for $\mathrm{Re} s > 1$. And as noted above, there is a real pole $s$ with $\mathrm{Re} s \geq 1$, and thus we also see that $s = 1$ is a pole.