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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.

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    $\begingroup$ Remarkably, the operation is associative. I was mistaken. $\endgroup$
    – AlvinL
    Commented Jan 29, 2023 at 11:22
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    $\begingroup$ I suggest starting by proving statements of the form $a,b<1\implies a@b<1$. At least that should let you prove that the even partial terms are $<1$ and the odd partial terms are $>1$. $\endgroup$
    – lulu
    Commented Jan 29, 2023 at 11:24
  • $\begingroup$ Ok, I will try. $\endgroup$
    – Maximax67
    Commented Jan 29, 2023 at 11:26
  • $\begingroup$ Let $x_3=2$ and for any $n\ge3,$ $x_{n+1}=\frac{n+x_n}{nx_n+1}.$ Then, $x_{n+1}-1=(1-x_n)\frac{n-1}{nx_n+1}.$ $\endgroup$ Commented Jan 29, 2023 at 13:12

3 Answers 3

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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 $$\lim_{n\to\infty}y_n=\lim_{+\infty}\tanh=1\quad\text{hence}\quad\lim_{n\to\infty}x_n=1.$$

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    $\begingroup$ +1 Very slick ${}{}$ $\endgroup$
    – FShrike
    Commented Jan 29, 2023 at 13:38
  • $\begingroup$ What technique do you use the solve the $y_n$ recurrence sequence? $\endgroup$
    – Ovi
    Commented Jan 29, 2023 at 19:32
  • $\begingroup$ I recognized the addition formula for $\tanh.$ $\endgroup$ Commented Jan 29, 2023 at 19:33
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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.

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  • $\begingroup$ Thank you. I understood the main idea. But could you please explain this in more detail? How to prove that subsequence of even partial terms is strictly increasing and subsequence of odd partial terms is strictly decreasing? Maybe this is a stupid question, but I can't figure out how to do it. $\endgroup$
    – Maximax67
    Commented Jan 29, 2023 at 12:11
  • $\begingroup$ @Maximax67 I explained odd partial terms, hope that makes it clearer. $\endgroup$
    – AlvinL
    Commented Jan 29, 2023 at 12:36
  • $\begingroup$ Thank you very much! $\endgroup$
    – Maximax67
    Commented Jan 29, 2023 at 12:37
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    $\begingroup$ While the even partial terms are strictly increasing and the odd partial terms are strictly decreasing, and both these subsequences are bounded by $1$, it's still necessary to prove that both converge to $1$ (but not two different limits away from $1$). $\endgroup$
    – peterwhy
    Commented Jan 29, 2023 at 16:16
  • $\begingroup$ @peterwhy My previous answer is redundant, I corrected the squeezing argument with a correct bound. $\endgroup$
    – AlvinL
    Commented Jan 31, 2023 at 9:43
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$\DeclareMathOperator{@}{\operatorname@}$ 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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