If $a_1=2$, $a_{n+1}=a_n+\frac{a_n}{a_n^2-1}$, show that $a_{2024}<64$

Question

If sequence $\{a_n\}$ is such that $a_1=2$ and $a_{n+1}=a_n+\frac{a_n}{a_n^2-1}$, show that $a_{2024}<64$.

First, I tried $a_{n+1}<a_n+\frac{a_n+1}{a_n^2-1}=a_n+\frac1{a_n-1}$. If $b_n=a_n-1$, then $b_{n+1}<b_n+\frac1{b_n}$. After squaring and summation, we get $$b_{n+1}^2<2n+b_1^2+\sum_{i=1}^n\frac1{b_i^2}.$$ We need to find a lower bound for $b_n$ ($a_n$). In fact, $a_{n+1}>a_n+\frac1{a_n}$, so squaring and summation gives $a_n\ge\sqrt{2n+2}$. Thus, $$b_{n+1}^2<2n+1+\sum_{i=1}^n\frac1{\left(\sqrt{2i+2}-1\right)^2}.$$ I checked that this bound is not enough.

Second, I tried $a_n=a_1+\sum\limits_{i=1}^{n-1}\frac{a_i}{a_i^2-1}$. Since $\frac x{x^2-1}$ is a decreasing function, we apply the above lower bound of $a_n$ again, but this is not enough as well.

Answer 1

By your work $$a_{2024}^2=a_1^2+\sum_{k=1}^{2023}\left(\frac{2a_k^2}{a_k^2-1}+\frac{a_k^2}{(a_k^2-1)^2}\right)=4+\sum_{k=1}^{2023}\frac{a_k^2(2a_k^2-1)}{(a_k^2-1)^2}=$$ $$=4050+\sum_{k=1}^{2023}\left(\frac{a_k^2(2a_k^2-1)}{(a_k^2-1)^2}-2\right)=4050+\sum_{k=1}^{2023}\frac{3a_k^2-2}{(a_k^2-1)^2}\leq$$ $$\leq4050+\sum_{k=1}^{2023}\frac{6k+4}{(2k+1)^2}<4050+\int\limits_0^{2023}\frac{6x+4}{(2x+1)^2}dx<4096.$$

Answer 2

$b_n = a_n^2$ satisfies $b_1 = 4$ and $$ b_{n+1} = b_n + \frac{b_n(2b_n-1)}{(b_n-1)^2} = b_n + 2 + \frac{3}{b_n-1} + \frac{1}{(b_n-1)^2} \, . $$ You already proved that $b_n \ge 2n+2 > 2n+1$, therefore is $$ b_{n+1} < b_n + 2 + \frac{3}{2n} + \frac{1}{4 n^2} \, . $$ It follows that $$ \begin{align} b_n &< 2n+2 + \frac 32 \sum_{k=1}^{n-1} \frac 1k + \frac 14 \sum_{k=1}^{n-1}\frac{1}{k^2} \\ &< 2n+2 + \frac 32 \bigl(1 + \ln(n)\bigr) + \frac 14 \frac{\pi^2}{6} \\ &< 2n + \ln(n) + 4 \, . \end{align} $$ In particular is $$ b_{2024} < 4052 + \frac 32 \ln(2024) \approx 4063.41925 < 4096 $$ and therefore $a_{2024} < 64$.

Answer 3

$b_n:=a_n^2$ satisfies $b_1=4,\quad b_{n+1}=\frac{b_n^3}{(b_n-1)^2}$ hence $c_n:=b_n-1$ satisfies$$c_1=3,\quad c_{n+1}=\frac{(c_n+1)^3}{c_n^2}-1=c_n+2+\frac{3c_n+1}{c_n^2}>c_n+2.$$ Therefore,$$c_n>3+2(n-1)=2n+1,$$ hence $$\begin{align}c_{2024}&<4049+\sum_{k=1}^{2023}\frac{6k+4}{(2k+1)^2} \\&<4095\\&=64^2-1.\end{align}$$

Answer 4

Consider the function $f(x) = x+ \frac{x}{x^2-1}$. Our sequence is $$a_n = f^n(2)$$ for $n\ge 1$. We have $f'(x) = \frac{x^2(x^2-3)}{(x^2-1)^2}$, so $f$ is strictly increasing on $[\sqrt{3}, \infty)$.

Consider also the function $g(t) = \sqrt{2 t + \sqrt{t}}$. We want to show that $f^n(2)< g(n)$ for $n$ large enough. It is enough to show that

1. $f^{N}(2) < g(N)$ for some low $N$
2. $f(g(t))< g(t+1)$

The first $N$ for which the above holds is $N=112$, since $f^{112}(2) =15.315\ldots$, while $g(112)=15.316\ldots$.

Now the inequality $2$ holds for $t\ge 8$, as one can check.

Now by induction one sees that $f^{n}(2) < g(n)$ for $n\ge 112$. In particular $f^{2024}(2) < g(2024)=63.97\ldots < 64$.

Note: We have $a_n \simeq \sqrt{2n}$. Indeed, for the sequence $b_n = \frac{1}{a_n}$ we have the recurrence $b_{n+1} = b_n(1-b_n^2)$, very similar to $b_{n+1} = \sin b_n$, and these kind of recurrences appear elsewhere on the site.