Let $n \in \Bbb{N}$ ( for me $0\notin \Bbb{N})$.
Find the limit as $n$ tends to $+ \infty$ of the following sequence $$\frac{\alpha_{n}}{n}$$ where $\alpha_{n}$ is the number of permutations of $\mathfrak{S}_{n}$ which all length cycles are less than $\frac{n}{2}$.
My try,
If $n$ is even we just need to break the set into two parts and notice that all hyperbolic subgroups of $\mathfrak{S}_{n}$ stabilizing both parts works, so we have $\alpha_n\ge (\frac{n}{2}!)^2$. Then the limit is $+ \infty$. How can I do for $n$ odd ?
An additional question
Can we find a 'simple' equivalent of $\alpha_{n}$?
Let's count the number of permutations with a cycle of length $\ge n/2$. It can only have one such cycle, for obvious reasons. First, pick a natural $n/2\le k\le n$. Next, pick a $k$-cycle from $S_n$, and finally the rest of the cycle decomposition can be determined by choosing any permutation on the set of elements in $\{1,\cdots,n\}$ not already present in our first chosen cycle.
Therefore,
$$\alpha_n=n!-\sum_{n/2\le k\le n} \frac{(n)_k}{k}(n-k)!=n!\left(1-\sum_{n/2\le k\le n}\frac{1}{k}\right).$$
Above, $(q)_k:=\overbrace{q(q-1)(q-2)\cdots}^k$ is the falling factorial. The number of sequences of length $k$ of elements of $\{1,\cdots,n\}$ with no repetitions is $(n)_k$, so we divide this by $k$ to get the total number of cycles in $S_n$ of length $k$. There are $n-k$ elements in $\{1,\cdots,n\}$ not present in a given $k$-cycle, which means our final choice is from $(n-k)!$ possible permutations. Clearly $(n)_k(n-k)!=n!$ so we used that to simplify.
Assuming the question was for $\alpha_n/n!$, we can use the estimate $H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+o(1)$ for the harmonic numbers to conclude that $\alpha_n/n!$ converges to $\color{Red}{1-\log 2}$.