Ttdfjt

$sin^2 theta$">

Where the modulus-like symbol actually denotes iteration of a radical function. Sorry for the messy work everyone- I am new to this stuff, and I literally just found this iteration formula like 15 minutes ago. $X[i]$ starts as being an arbitrary number greater than $n!/(n-2)!$ which I should have mentioned earlier but didn't have space to (apologies).
So the business is: you start with a number e.g. 1, then if you want to find say $50$, you compute $sqrt(1 + 50!/48!)$ then replace $1$ on your calculator with the result you just got a.k.a. $Ans$ etc. and keep iterating till you get (at a surprisingly fast convergence rate!) the very number you started with, n.

I appreciate this might not be the most interesting insight or $n!/(n-2)!$ is probably just me not realising that it could also be presented as $n(n-1) = n^2-n$. But still, please don't mass-downvote it all, remember it's a 15-year old person who figured this out...

You can do this formally by just looking at
$$sqrtn!/(n-2)!+sqrtn!/(n-2)! + dots$$
and so on to infinity. Suppose this has a limit. Then we'd expect that
$$L=sqrtn!/(n-2)! + L$$
This gives
$$L^2-L=n!/(n-2)!=n^2-n$$
which has the obvious solution of $L=n$.


Analytically, this could be done with some sort of fixpoint theorem (such as the Banach fixed-point theorem), setting
$$f(x)=sqrtn!/(n-2)! + x$$
and using it to show that $lim_mto inftyunderbracef(cdots f(x)cdots)_mtext times$ is the solution to the fixpoint equation
$$f(L)=L.$$

So you're saying that the sequence
$$X_0=1$$
$$X_k+1=sqrtX_k+fracn!(n-2)!$$
tends to towards $n$. Firstly notice that
$$fracn!(n-2)!=n(n-1)=n^2-n$$
Hence your sequence is equivalent to
$$X_k+1=sqrtX_k+n^2-n$$
Then note that for $0le X_klt n$ we have
$$X_kltsqrtX_k+n^2-nlt n$$
So your sequence is strictly increasing and bounded above. Also note that
$$sqrtn+n^2-n=sqrtn^2=n$$
Hence the sequence has a stationary point at $X_k=n$. This is sufficient to prove that for a given initial value $0le X_0le n$ the sequence tends to $n$. In other words
$$lim_ktoinftyX_k=n$$
In fact one can prove that this sequence tends to $n$ for any initial value by also using the fact that $sqrtX_k+n^2-nlt X_k$ when $X_kgt n$.