Given $(a_n)$ such that $a_1 \in (0,1)$ and $a_{n+1}=a_n+(\frac{a_n}{n})^2$. Prove that $(a_n)$ has a finite limit. [duplicate]

Given $(a_n)$ such that $a_1 \in (0,1)$ and $a_{n+1}=a_n+(\frac{a_n}{n})^2$. Prove that $(a_n)$ has a finite limit. Clearly $a_n$ are increasing. Also, $$\frac{1}{a_{n+1}}=\frac{1}{a_n\left(1+\frac{a_n}{n^2}\right)}=\frac{1}{a_n}-\frac{\frac{1}{n^2}}{1+\frac{a_n}{n^2}}$$

From this how to proceed further? Any idea will be appreciated

EDIT. I realized that this problem has been solved (even by myself) several times in the past; see the links above. A bound of the form

$$ \frac{x}{1-x} \leq \lim a_n \leq \frac{x}{1-\sqrt{x}} $$

is also proved there.

Let $x = a_1 \in (0, 1)$.

Claim 1. We have $a_n \leq nx$ for all $n \geq 1$.

Indeed, the base case ($n=1$) is trivial. Next, if $ a_n \leq nx$, then

$$ a_{n+1} \leq nx + \left( \frac{nx}{n} \right)^2 = nx + x^2 \leq (n+1) x. $$

So by the mathematical induction, the claim follows.

Claim 2. We have $a_n \leq Cn^x$ for some constant $C = C(x) \in (0, \infty)$.

Indeed,

$$ a_{n+1} = a_n \left( 1 + \frac{a_n}{n^2} \right) \leq a_n \left( 1 + \frac{x}{n} \right) \leq a_n e^{\frac{x}{n}} $$

where we utilized the inequality $1 + t \leq e^t$ in the last step. From this, we get

$$ a_n \leq x e^{x H_{n-1}}, $$

where $H_n = \sum_{k=1}^{n} \frac{1}{k}$ is the harmonic number. Then using $H_n = \log n + \mathcal{O}(1)$, the desired claim follows. (In fact, using the inequality $H_{n-1} \leq 1+ \log n$, we may choose $C = x e^{x}$.)

Claim 3. $(a_n)$ converges.

Let $C$ be the constant as in the previous claim. Then

$$ a_{n+1} = a_n \left( 1 + \frac{a_n}{n^2} \right) \leq a_n \left( 1 + \frac{C}{n^{2-x}} \right) \leq a_n \exp\left(\frac{C}{n^{2-x}}\right), $$

and so,

$$ a_n \leq x \exp\left( \sum_{k=1}^{n-1} \frac{C}{k^{2-x}} \right) \leq x e^{C \zeta(2-x)}. $$

This proves that $(a_n)$ is bounded. Since $(a_n)$ is increasing, we conclude that $(a_n)$ converges.

Addendum. A numerical simulation suggests that $a_{\infty}$ as a function of $x = a_1$ satisfies

$$ a_{\infty}(x) \leq \frac{2x}{1-x} $$

for all $x \in (0, 1)$, although I could not prove this. On the other hand, we can obtain a lower bound of the same order via relatively simple argument:

Argument 1. From the recurrence relation and Claim 1,

\begin{align*} \frac{1}{x} - \frac{1}{a_{\infty}(x)} = \sum_{n=1}^{\infty} \frac{1}{n^2 + a_n(x)} \geq \sum_{n=1}^{\infty} \frac{1}{n^2 + nx} = \sum_{n=1}^{\infty} \frac{1}{x} \left( \frac{1}{n} - \frac{1}{n+x} \right) = \frac{H_x}{x}, \end{align*}

where $H_x = \int_{0}^{1} \frac{1-t^x}{1-t} \, \mathrm{d}t$ is the harmonic number. Rearranging,

$$ a_{\infty}(x) \geq \frac{x}{1 - H_x} \sim \frac{1}{1-x} \quad \text{as} \quad x \to 1^-. $$

Argument 2. By the principle of mathematical induction, it is easy to show that

$$ a_n \geq x + x^2 + \cdots + x^n = x \cdot \frac{1-x^n}{1-x}. $$

Indeed, the claim is trivial for $n = 1$, and if it is true for $n$, then by the AM-GM inequality, $a_n \geq n x^{(n+1)/2}$, and so,

\begin{align*} a_{n+1} &\geq (x + x^2 + \cdots + x^n) + \left( \frac{nx^{(n+1)/2}}{n} \right)^2 \\ &= x + x^2 + \cdots + x^n + x^{n+1}. \end{align*}

This establishes the inductive step and hence proves the desired claim.