Take $k,N\in \mathbb N$ and $s\in \mathbb C$ with real part $\sigma \in [1-\delta ,1]$ for some small fixed $\delta $. In its simplest form my question is how do I sum $$\sum _{l\geq 0}\frac {d_k(p^{l+N})}{p^{ls}}?$$ A longer version is: I have an Euler product that looks something like $$\prod _{p^N||n}\left (1+\frac {p^{-\sigma }\sum _{l\geq 0}d_k(p^{N+l})/p^{ls}}{\sum _{l\geq 0}d_k(p^{N+l})/p^{ls}+d_{k}(p^N)N/k}\right )$$ and I think I want each factor to be $\asymp 1$ but I'm not sure what stops the denominator from blowing up for small primes. If $k=2$ then each sum is $$\frac {1}{(1-p^{-s})^2}+\frac {N}{1-p^{-s}}$$ so the quotient is $$p^{-\sigma }\frac {1+N(1-p^{-s})}{1+N(1-p^{-s})+(N+1)N(1-p^{-s})^2/2}$$ which I hope I can think of as $$\approx p^{-\sigma }\frac {1+N}{1+N+N^2/2}$$ so I'm ok in the case $k=2$. But for general $k$ I'm not sure - I think part of my difficulty is I'm not really sure what to make of $$\mathcal S:=\sum _{l\geq 0}\frac {d_k(p^{l+N})}{p^{ls}}=\sum _{l\geq 0}{{l+N+k-1}\choose {k-1}}\frac {1}{p^{ls}}$$ which for $k=2$ I could calculate "by hand" but I get a bit lost for higher $k$.
If we write $x=p^{-s}$, $K=k-1$, $M=N+K$ then by https://en.wikipedia.org/wiki/Binomial_coefficient#Generating_functions and then Taylor's theorem, there is $\eta \in (0,|x|)$ such that $$\mathcal S=x^{-M}\sum _{l\geq M}{{l}\choose {K}}x^l=x^{-M}\left (\underbrace {\frac {x^K}{(1-x)^{K+1}}}_{=:f(x)}-\sum _{0\leq l\leq M-1}...\right )=\frac {f^{(M)}(\eta )}{M!}$$ so (if I've differentiated and tidied up correctly) $$(1-\eta )^{K+1+M}\mathcal S=\frac {1}{M!}\sum _{a+b=M\atop {a\leq K}}n\eta ^{K-a}(1-\eta )^{M-b}$$ where $$n={{M}\choose {a,b}}\frac {(K+b)!}{(K-a)!}.$$ Taking only the two end terms of this sum (just to try and have an idea of what happens) this is $$\approx n_1\eta ^K+n_2(1-\eta )^K$$ where $$n_1=\frac {(K+M)!}{M!K!}=\frac {(N+K+1)\cdot \cdot \cdot (N+2K)}{K!}$$ $$n_2=\frac {M!}{K!(M-K)!}=\frac {(N+1)\cdot \cdot \cdot (N+K)}{K!}=d_k(p^N)$$ so that the quotient in my Euler product is $$\frac {\mathcal S}{\mathcal S+d_k(p^N)N/k}=\frac {n_1\eta ^K+n_2(1-\eta )^K}{n_1\eta ^K+n_2(1-\eta )^K+(1-\eta )^{K+1+M}n_2N/k}\approx \frac {n_1\eta ^K+n_2}{n_1\eta ^K+n_2(1+N/k)}$$ but again I'm not sure if I can control the denominator here.
Mathematica says that $$\sum _{l\geq 0}{{l+N+k-1}\choose {k-1}}\frac {1}{p^{ls}} = \binom{N + k-1}{k-1} {}_2F_1(1, N + k, N+1; p^{-s}),$$ where ${}_2F_1$ is the hypergeometric function.
From numerical calculations, the function ${}_2F_1(1, N + k, N+1; x)$ with these integer parameters appears to always be a rational function with denominator $(x-1)^k$ and numerator a polynomial of degree $k-1$ in $x$. Indeed, the pattern appears to be $$ \binom{N + k-1}{k-1} {}_2F_1(1, N + k, N+1; x) = \sum_{j=0}^{k-1} \binom{N-1+j}j (1-x)^{j-k}. $$