Expectation of inverse of sum of iid random variables - approximation I am working with the set of i.i.d discrete random variables $\{\zeta_1, \ldots, \zeta_n\}$. Each one of them can take either of the $m$ values $\{z_1, \ldots, z_m\}$ with corresponding probabilities $\{p_1, \ldots , p_m\}$. 
I am trying to understand when I can apply the following approximation for the expectation (which I believe to be the first-order one):
\begin{equation}
\mathbb{E}\left[\min\left(A, \frac{1}{\sum_{k=1}^n \mathrm{I}[\zeta_k\geq \bar{z}]}\right)\right] \approx \min\left(A, \frac{1}{\mathbb{E}\left[\sum_{k=1}^n \mathrm{I}[\zeta_k\geq \bar{z}]\right]}\right)
\end{equation}
Here $A$ and $\bar{z}$-- are some constants and an indicator function $\mathrm{I}[\zeta_k\geq \bar{z}]=1$ if $\zeta_k\geq \bar{z}$ and $0$ otherwise.
I am working in the regime $n\to\infty$ and the question is whether the approximation above is good enough under this condition? (if it is possible to judge, though)
I checked the following link, however I am not sure how to make use of $\mathcal{L}_{X(t)}^n$ under large $n$ in the case of discrete distribution.
Thank you in advance for any help.

Let $p=P(\zeta\ge z)$. Then as $n\to\infty$ $$LHS\approx E(\frac 1 {B(n,p)};B(n,p)\ge 1)\approx \frac 1 {np}+\frac {np(1-p)}{(np)^3}=\frac 1 {np}+\frac {1-p}{(np)^2}=\frac 1{np}(1+\frac{1-p}{np})$$
while $RHS=\frac 1 {np}$. Hence $LHS\sim RHS$.
