I am new to this forum and I hope there is someone who might be able to help me with this limit. For a natural $n$, and with $\alpha$ a positive constant, denote with $p:=\alpha/\sqrt{n}$. Consider then the sum $$L(n) = \sum_{k=0}^n {n\choose k} p^k (1-p)^{n-k} (1-p^2)^{k(k-1)/2}$$ I am interested in determining $L:=\lim_{n\rightarrow\infty} L(n)$.
The setting can be sketched as follows: Consider an experiment in which we run $n$ times a Bernoully lottery with probability of success$=p$, and then, if there were $k$ successess, we run ${k\choose 2}$ times a second bernoully lottery with probability of success $p^2$. Each success of the second lottery, is a success of the whole experiment. Then, $L(n)$ should represent the probability that there were no successes of the whole experiment. The ideal would be if the limit is $0$, but another equally good result for me would be to prove, as I believe, that $$\frac{1}{\sqrt{e^{\alpha^4}}}$$ is an upper bound to the limit. Thanks in advance for any help that any of you could provide.
Indeed we have $L = e^{-\alpha^4/2}$.
Proof. Let $S_n \sim \text{Binomial}(n, \alpha/\sqrt{n})$. Then $L(n)$ can be written as
$$ L(n) = \mathbf{E}\Bigl[ (1 - \alpha^2/n)^{\binom{S_n}{2}} \Bigr]. $$
Now fix $\delta \in (0, \frac{1}{4})$, and define $a_n = \alpha\sqrt{n} - n^{1/4+\delta}$ and $b_n = \alpha\sqrt{n} + n^{1/4+\delta}$. Also, consider the event $A_n = \{ S_n \in [a_n, b_n] \}$. By the Chebyshev's inequality,
$$ \mathbf{P}(A_n^c) \leq \frac{\mathbf{Var}(S_n)}{n^{\frac{1}{2}+2\delta}} \leq \frac{C}{n^{2\delta}} $$
for some absolute constant $C > 0$ and hence $\mathbf{P}(A_n^c) \to 0$ as $n \to \infty$. Then by noting that $k \mapsto (1-p^2)^{\binom{k}{2}}$ is decreasing in $k$, we get
$$ (1 - \alpha^2/n)^{\binom{b_n}{2}} \mathbf{P}(A_n) \leq L(n) \leq (1 - \alpha^2/n)^{\binom{a_n}{2}} \mathbf{P}(A_n) + \mathbf{P}(A_n^c), $$
and it is easy to check that both the lower and upper bound converge to $e^{-\alpha^4/2}$, proving OP's claim.