Fix $\varepsilon \in (0, 1), \lambda >0$, and a positive integer $n$. I'm interested in upper bound the quantity $$ S_n := \sum_{k=1}^n e^{-\lambda} \frac{\lambda^k}{k!} \frac{1}{(k \varepsilon)^k} $$ in terms of $\varepsilon,\lambda,n$. The term $e^{-\lambda} \frac{\lambda^k}{k!}$ is the probability mass from Poisson distribution with parameter $\lambda$. The term $\frac{1}{(k \varepsilon)^k}$ can be considered the loss associated to $k$. So $S_n$ can be seen as the average loss up to $n$. Of course, we have a trivial upper bound $$ S_n \le \sum_{k=1}^\infty e^{-\lambda} \frac{\lambda^k}{k!} \frac{1}{\varepsilon^k} = \exp \left (-\lambda + \frac{\lambda}{\varepsilon} \right ). $$
Could you elaborate on some techniques to have a tight upper boud of $S_n$? Any reference is appreciated!
We can use the asymptotically sharp inequality $$ \Gamma\! \left( {k + \tfrac{1}{2}} \right) \le k^k {\rm e}^{ - k} \sqrt {2\pi } $$ valid for all $k\ge 1$ (see, e.g., this paper). This yields \begin{align*} \sum\limits_{k = 1}^n {{\rm e}^{ - \lambda } \frac{{\lambda ^k }}{{k!}}\frac{1}{{(\varepsilon k)^k }}} & \le \sqrt {2\pi } {\rm e}^{ - \lambda } \sum\limits_{k = 1}^n {\frac{1}{{k!\Gamma\! \left( {k + \frac{1}{2}} \right)}}\left( {\frac{\lambda }{{{\rm e}\varepsilon }}} \right)^k } \\ & = \sqrt 2 {\rm e}^{ - \lambda } \sum\limits_{k = 1}^n {\frac{1}{{(2k)!}}\left( {\frac{{4\lambda }}{{{\rm e}\varepsilon }}} \right)^k } \\ & \le \sqrt 2 {\rm e}^{ - \lambda } \sum\limits_{k = 1}^\infty {\frac{1}{{(2k)!}}\left( {\frac{{4\lambda }}{{{\rm e}\varepsilon }}} \right)^k } \\ &= \sqrt 2 {\rm e}^{ - \lambda } \left( {\cosh \left( {2\sqrt {\frac{\lambda }{{{\rm e}\varepsilon }}} } \right) - 1} \right). \end{align*} You may obtain better bounds by estimating $$ 2\sum\limits_{k = 1}^n {\frac{{z^{2k} }}{{(2k)!}}} = {\rm e}^z Q(2n + 1,z) + {\rm e}^{ - z} Q(2n + 1, - z) - 2 $$ in a different manner (here $Q$ is the normalised incomplete gamma function).