Please explain how to solve limit. I know the answer but how to explain it? Problem:
$a@b = \frac{a+b}{ab+1}$. Solve limit:
$\lim_{n \to \infty}(2@3@...@n)$.
I've tried to solve this problem by just calculating:
$$2 @ 3 = 0.714$$
$$2 @ 3 @ 4 = 1.222$$
$$2 @ 3 @ 4 @ 5 = 0.875$$
$$2 @ 3 @ 4 @ 5 @ 6 = 1.1$$
I found the pattern. The first number is less than 1, then the next is greater than 1, the next is less than 1, and so on. So the limit must be 1. But how to explain it mathematically?
I've tried to transform this:
$$(n-1)@n = \frac{2n-1}{n^2-n+1}$$
$$n@(n+1) = \frac{2n + 1}{n^2+n+1}$$
But it didn't help me to understand the method how to solve it. I think there should be a simple idea, which I don't see. I appreciate all hints.
 A: It holds that $\lim x_n = x$ if and only if every subsequence of $x_n$ contains, in turn, a subsequence that converges to $x$.
Check that the subsequence of even partial terms is strictly increasing and of odd partial terms is strictly decreasing.
Take any subsequence of the initial sequence. If it contains odd partial terms infinitely often, then it contains a subsequence converging to $1$. Otherwise, if it eventually contains only even partial terms, it again contains a subsequence converging to $1$. Thus, the initial limit is $1$.

Let's examine odd partial terms, say. It is readily verified that $@$ is associative and
$$ a@b@c = \frac{a+b+c+abc}{ab+bc+ca+1}.$$
My pre-edit answer is redundant. Also, the upper bound of $1+\frac{1}{n}$ does not hold, I miscalculated. We'll have to be more lenient with the bound. We'll accept odd partial terms being $>1$ as given.
It suffices to check that
$$ 2@3@\ldots @2n \leqslant 1+\frac{1}{\sqrt{n}},\quad n\in\mathbb N $$
Base case holds. Suppose $A:= 2@3@\ldots @2n \leqslant 1+\frac{1}{\sqrt{n}}$ for some $n$. Then
$$
\begin{align*}
 A@(2n+1)@(2n+2) &= \frac{A+(2n+1)+(2n+2)+A(2n+1)(2n+2)}{A(2n+1)+(2n+1)(2n+2)+A(2n+2)+1} \\
&\leqslant \frac{A+(2n+1)+(2n+2)+A(2n+1)(2n+2)}{(2n+1)+(2n+1)(2n+2)+(2n+2)+1} \\
&= \frac{(4n+3) + A((2n+1)(2n+2)+1)}{(2n+1)+(2n+1)(2n+2)+(2n+2)+1} \\
&\leqslant \frac{(4n+3) + (1+\frac{1}{\sqrt{n}})((2n+1)(2n+2)+1)}{(2n+1)+(2n+1)(2n+2)+(2n+2)+1} \\
&= \frac{4\sqrt{n^3} + 4n^2 + 10n + 6\sqrt{n} + \frac{3}{\sqrt{n}}+6}{4n^2 + 10n+6} \\
&\overset{?}\leqslant 1+\frac{1}{\sqrt{n+1}}
\end{align*}
$$
The last inequality is a matter of direct verification. It suffices to check
$$\frac{4\sqrt{n^3}+6\sqrt{n}+\frac{3}{\sqrt{n}}}{4n^2+10n+6} \leqslant \frac{1}{\sqrt{n+1}}. $$
Note that
$$ \frac{4n^2+6n+3}{\sqrt{\frac{n}{n+1}}} \leqslant 4n^2+10n+6 \Leftrightarrow 8n(n+1)(4n+3)\geqslant 3, $$
the right hand statement is evidently true. Even partial terms can be tackled analogously.
A: $\DeclareMathOperator{@}{\operatorname@}$
As noted in a comment, the $\@$ operator is associative.
By expanding $(a\@ b\@c)$ and $(a\@b\@c\@d)$:
$$\begin{align*}
a\@b &= \frac{a+b}{ab+1}\\
(a\@b)\@c &= \frac{\frac{a+b}{ab+1}+c}{\frac{a+b}{ab+1}c+1}= \frac{a+b+c(ab+1)}{(a+b)c + ab+1} = \frac{abc + a+b+c}{ab+ac+bc+1}\\
(a\@b\@c)\@d &= \frac{\frac{abc + a+b+c}{ab+ac+bc+1}+d}{\frac{abc + a+b+c}{ab+ac+bc+1}d+1}\\
&= \frac{abc + a+b+c + (ab+ac+bc+1)d}{(abc + a+b+c)d + ab+ac+bc+1}\\
&= \frac{abc+abd+acd+bcd + a+b+c+d}{abcd + ab+ac+ad+bc+bd+cd + 1}
\end{align*}$$
Note that the numerators and denominators seem to be sums of the elementary symmetric polynomials of $2$, $3$ or $4$ variables. Define a polynomial $f_n(x)$ for $n\ge 2$,
$$\begin{align*}
f_n(x) &= (2x+1)(3x+1)\cdots (nx+1)\\
&= (2\cdot 3\cdots n)x^{n-1} + \cdots + (2+3+\cdots + n) x + 1
\end{align*}$$
(Note that, as in the question, here $n$ is in the largest operand, not the number of operands or partial terms.)
Then the denominators of the partial terms seem to be the sum of the coefficients of $x^0$ and every second term; the numerators of the partial terms seem to be the sum of the coefficients of $x^1$ and every second term.
Claim that
$$\begin{align*}
2 \@3\@\cdots \@n
&= \frac{\frac12\left[f_n(1)- f_n(-1)\right]}{\frac12\left[f_n(1)+ f_n(-1)\right]}\\
&= \frac{3\cdot4\cdots(n+1) - (-1)^{n-1}\cdot 1\cdot2\cdots(n-1)}{3\cdot4\cdots(n+1) + (-1)^{n-1}\cdot 1\cdot2\cdots(n-1)}\\
&= \frac{(n-1)!\cdot \left[\frac{n(n+1)}2+(-1)^n\right]}{(n-1)!\cdot \left[\frac{n(n+1)}2-(-1)^n\right]}\\
&= \frac{n(n+1)+2(-1)^n}{n(n+1)-2(-1)^n}
\end{align*}$$
For the base case $n=2$,
$$\begin{align*}
LHS &= 2\\
RHS &= \frac{2\cdot3+2\cdot(-1)^2}{2\cdot3-2\cdot(-1)^2} = \frac{8}4 = 2
\end{align*}$$
Assume for some integer $k\ge 2$ that the claim is true:
$$2\@3\@\cdots \@k = \frac{k(k+1)+2(-1)^k}{k(k+1)-2(-1)^k}$$
Then for the $n=k+1$ case,
$$\begin{align*}
LHS &= 2\@3\@\cdots\@k\@(k+1)\\
&= \left[\frac{k(k+1)+2(-1)^k}{k(k+1)-2(-1)^k}\right] \@ (k+1)\\
&= \frac{\left[\frac{k(k+1)+2(-1)^k}{k(k+1)-2(-1)^k}\right]+(k+1)}{\left[\frac{k(k+1)+2(-1)^k}{k(k+1)-2(-1)^k}\right](k+1)+1}\\
&= \frac{k(k+1)+2(-1)^k+\left[k(k+1)-2(-1)^k\right](k+1)}{\left[k(k+1)+2(-1)^k\right](k+1)+k(k+1)-2(-1)^k}\\
&= \frac{k(k+1)(k+2)-2(-1)^kk}{k(k+1)(k+2) + 2(-1)^kk}\\
&= \frac{(k+1)(k+2)+2(-1)^{k+1}}{(k+1)(k+2)-2(-1)^{k+1}}\\
&= RHS
\end{align*}$$
So by induction, for integers $n\ge 2$,
$$\begin{align*}
2 \@3\@\cdots \@n
&= \frac{n(n+1)+2(-1)^n}{n(n+1)-2(-1)^n}
\end{align*}$$
Taking the limit when $n\to \infty$,
$$\begin{align*}
\lim_{n\to\infty}(2 \@3\@\cdots \@n)
&= \lim_{n\to\infty}\frac{n(n+1)+2(-1)^n}{n(n+1)-2(-1)^n}\\
&= \lim_{n\to\infty}\frac{1+\frac{2(-1)^n}{n(n+1)}}{1-\frac{2(-1)^n}{n(n+1)}}\\
&= \frac{1+0}{1-0}\\
&= 1
\end{align*}$$
A: Let
$$x_n=2@3@...@n$$
and
$$y_2=\frac1{x_2},\quad y_3=x_3,\quad y_4=\frac1{x_4},\quad y_5=x_5\dots.$$
Then,
$$y_n=\frac{y_{n-1}+\frac1n}{1+\frac1ny_{n-1}}$$
hence
$$y_n=\tanh\sum_{k=2}^n\operatorname{artanh}\frac1k$$
so $(y_n)$ is positive, increasing, and $<1,$ whence the behaviour of $(x_n)$. Moreover, from $\operatorname{artanh}x\sim_{x\to0}x$ and $\sum_{k\ge2}\frac1k=+\infty$ we deduce $\sum_{k\ge2}\operatorname{artanh}\frac1k=+\infty,$ so
$$\lim_{n\to\infty}y_n=\lim_{+\infty}\tanh=1\quad\text{hence}\quad\lim_{n\to\infty}x_n=1.$$
