A good upper bound of $S_n := \sum_{k=1}^n e^{-\lambda} \frac{\lambda^k}{k!} \frac{1}{(k \varepsilon)^k}$ in terms of $\varepsilon,\lambda,n$ 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!

 A: 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).
A: Fact 1: $x^{-x} \le a^{-x} \mathrm{e}^{-x + a}$ for all $x, a > 0$.
(Proof: Taking logarithm on both sides, letting $u = \frac{a}{x} > 0$,
it is equivalent to $\ln u \le u - 1$ which is true (easy).)
By Fact 1, we have, for all $a > 0$,
\begin{align*}
 \sum_{k=1}^n \mathrm{e}^{-\lambda} \frac{\lambda^k}{k!} \frac{1}{(k \varepsilon)^k} &\le \sum_{k=1}^n \mathrm{e}^{-\lambda} \frac{\lambda^k}{k!} \frac{1}{ \varepsilon^k} \cdot a^{-k} \mathrm{e}^{-k + a}\\[6pt]
 &= \mathrm{e}^{-\lambda + a} \sum_{k=1}^n \frac{1}{k!}\left(\frac{\lambda}{\mathrm{e} a \varepsilon }\right)^k\\[6pt]
 &\le \mathrm{e}^{-\lambda + a} \sum_{k=1}^\infty \frac{1}{k!}\left(\frac{\lambda}{\mathrm{e} a \varepsilon }\right)^k\\[6pt]
 &= \mathrm{e}^{-\lambda + a + \lambda/(\mathrm{e} a \varepsilon) } - \mathrm{e}^{-\lambda + a}.
\end{align*}
Thus, we have
$$\sum_{k=1}^n \mathrm{e}^{-\lambda} \frac{\lambda^k}{k!} \frac{1}{(k \varepsilon)^k} \le \min_{a>0} \left(\mathrm{e}^{-\lambda + a + \lambda/(\mathrm{e} a \varepsilon) } - \mathrm{e}^{-\lambda + a}\right).$$
Since the RHS does not admit a simple solution,
we may let e.g. $a = 1$ to get
$$\sum_{k=1}^n \mathrm{e}^{-\lambda} \frac{\lambda^k}{k!} \frac{1}{(k \varepsilon)^k} \le
\mathrm{e}^{-\lambda + 1 + \lambda/(\mathrm{e}  \varepsilon) } - \mathrm{e}^{-\lambda + 1}.$$
