<s>[INST] 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.
 [/INST] $\DeclareMathOperator{@}{\operatorname@}$
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 $n\to \infty$,
$$\begin{align*}
\lim_{n\to\infty}(2 \@3\@\cdots \@n)
&= \lim_{n\to\infty}\frac{n(n+1)+2(-1)^n}{n(n+1)-2(-1)^n}\\
&= \lim_{n\to\infty}\frac{1+\frac{2(-1)^n}{n(n+1)}}{1-\frac{2(-1)^n}{n(n+1)}}\\
&= \frac{1+0}{1-0}\\
&= 1
\end{align*}$$
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