Let $X_1,X_2,\ldots,X_n$ be a sequence of $n$ i.i.d. chi-squared random variables with $k$ degrees of freedom, and denote by $X_\max$ the maximum of this sequence. Furthermore, let $k=\omega(1)$ increase, and $n$ be an increasing function of $k$, $n=f(k)$, where $f(k)$ is not increasing too fast, i.e. $\log(n)=o(k)$. For example, $n=k^d$ for some $d>0$. This is very similar to a scenario in my previous question.

Now, let

$$S(n)=\sum_{i=1}^n\exp\left[g(n)\frac{\sqrt{\log n}}{\sqrt{k}}(X_i-X_\max)\right]$$

where $g(n)=o(1)$ is a positive but slowly decreasing function (which we can pick arbitrarily, as long as it's decreasing).

I am wondering about the asymptotic behavior of the sum $S(n)$ as $n\rightarrow\infty$. Specifically, I wonder how does $S(n)$ grow in terms $n$, $k$ and $g(n)$? I.e., how much do the terms other than maximum matter as $n$ gets large? I'll be happy with an in-distribution convergence...

**What I've done**

Let's denote the terms in the sum by $Y_i=\exp\left[g(n)\frac{\sqrt{\log n}}{\sqrt{k}}(X_i-X_\max)\right]$. Clearly, each $Y_i$ is bounded: $0\leq Y_i\leq 1$. Now, using the fact that $\frac{X_\max-k}{\sqrt{2k\log n}}\rightarrow 1$ almost surely from the answer to my previous question, one can show that each $Y_i$ *individually* has low probability of being close to unity (i.e. $Y_\max$). So that means that $S(n)$ is not growing linearly with $n$. However, one can also show that $P(Y_i\leq \delta)\rightarrow 0$ for any $\delta=o(1/n^c)$, $c>0$. So $S(n)$ is growing with $n$, but how? Any hints/tips/suggestions would be appreciated...

**Note**

I substantially revised this question since I figured out the actual question that I wanted to ask. @ofer zeitouni's comment refers to the previous version of this question...