Approximation the sum $\sum\limits_{n=0}^\infty (c+n)^{k-1} e^{-\frac{(c+n)^k}{2*a}}$

I would like to find lower and upper bounds on the following sum

\begin{align}\sum_{n=0}^\infty (c+n)^{k-1} e^{-\frac{(c+n)^k}{2*a}}\end{align}where $c,a>0$ and $k>1$.

Note that if we could approximate this sum with an integral then \begin{align}\sum_{n=0}^\infty (c+n)^{k-1} e^{-\frac{(c+n)^k}{2*a}} \approx \int _{n=0}^\infty (c+n)^{k-1} e^{-\frac{(c+n)^k}{2*a}} dn= -\frac{2 a}{k} e^{-\frac{(c+n)^k}{2*a}} |_0^\infty=\frac{2 a}{k} e^{-\frac{c^k}{2*a}}\end{align}

then one might guess that the bounds are \begin{align}\frac{2 a}{k} e^{-\frac{c^k}{2*a}}+c_1 \le\sum_{n=0}^\infty (c+n)^{k-1} e^{-\frac{(c+n)^k}{2*a}} \le \frac{2 a}{k} e^{-\frac{c^k}{2*a}}+c_2\end{align}

My question is how can this be done or is there a better method to give lower and upper bounds?

I also know that there are bounds of the form\begin{align}\int_0^\infty f(x) dx \le \sum_{n=0}^\infty f(n) \le f(0) + \int_0^\infty f(x) dx\end{align}

but they only hold if $f(x)$ is monotone decreasing which is not the case here.

Thanks.