Prove the existence of limit of $x_{n+1}=x_n+\frac{x_n^2}{n^2}$ The problem is: Let $\{x_n\}$ be a sequence such that $0<x_1<1$ and $x_{n+1}=x_n+\dfrac{x_n^2}{n^2}$. Prove that there exists the limit of $\{x_n\}$.
It is easy to show that $x_n$ is increasing, but I cannot prove it is bounded to show the existing of limit. Anyone have any ideas?
 A: Since $0\lt x_1\lt1$, the relation
$$
x_{n+1}=x_n+\frac{x_n^2}{n^2}\tag{1}
$$
implies not only that $x_n$ is increasing, but also inductively that
$$
0\lt x_n\lt n\tag{2}
$$
Equation $(1)$ implies
$$
\begin{align}
\left(\frac1{x_{n+1}}-\frac1{n+1}\right)-\left(\frac1{x_n}-\frac1n\right)
&=\left(\frac1n-\frac1{n+1}\right)-\left(\frac{n^2+x_n}{n^2x_n+x_n^2}-\frac{n^2}{n^2x_n+x_n^2}\right)\\
&=\frac1{n^2+n}-\frac1{n^2+x_n}\\
&=-\frac{n-x_n}{(n^2+n)(n^2+x_n)}\\
&\ge-\frac{n-x_n}{n^3(n+1)}\\
&\ge-\frac{n-x_n}{x_nn^2(n+1)}\\
&=-\frac1{n(n+1)}\left(\frac1{x_n}-\frac1n\right)\tag{3}
\end{align}
$$
Therefore,
$$
\left(\frac1{x_{n+1}}-\frac1{n+1}\right)
\ge\left(1-\frac1{n(n+1)}\right)\left(\frac1{x_n}-\frac1n\right)\tag{4}
$$
and since
$$
\prod_{n=1}^\infty\left(1-\frac1{n(n+1)}\right)=\frac1\pi\sin\left(\frac\pi\phi\right)\tag{5}
$$
we have
$$
\frac1{x_n}-\frac1n\ge\frac1\pi\sin\left(\frac\pi\phi\right)\left(\frac1{x_1}-1\right)\tag{6}
$$
which means
$$
\bbox[5px,border:2px solid #C0A000]{x_n\le\pi\csc\left(\frac\pi\phi\right)\frac{x_1}{1-x_1}}\tag{7}
$$
where $\pi\csc\left(\frac\pi\phi\right)=3.3706903036$.
Therefore, since $x_n$ is increasing and bounded above, the limit exists and is bounded by $(7)$.

Motivational Note
I looked at
$$
x_{n+1}-x_n=\frac{x_n^2}{n^2}\tag{8}
$$
as representative of
$$
\frac{\mathrm{d}x}{\mathrm{d}n}=\frac{x^2}{n^2}\tag{9}
$$
whose solution is
$$
\frac1x-\frac1n=C\tag{10}
$$
So I considered $\frac1{x_n}-\frac1n$ which lead to $(3)$, $(4)$, $(6)$ and $(7)$.

Derivation of $\boldsymbol{(5)}$
$$
\begin{align}
\prod_{k=1}^n\left(1-\frac1{k(k+1)}\right)
&=\prod_{k=1}^n\frac{(k+1-\phi)(k+\phi)}{k(k+1)}\tag{11}\\
&=\underbrace{\frac{\Gamma(n+2-\phi)}{\Gamma(2-\phi)}}_{\prod(k+1-\phi)}
\underbrace{\frac{\Gamma(n+1+\phi)}{\Gamma(1+\phi)}}_{\prod(k+\phi)}
\underbrace{\frac{\Gamma(1)}{\Gamma(n+1)}}_{\prod\frac1k}
\underbrace{\frac{\Gamma(2)}{\Gamma(n+2)}}_{\prod\frac1{k+1}}\tag{12}\\
&\to\frac1{\Gamma(2-\phi)\Gamma(1+\phi)}\tag{13}\\[6pt]
&=\frac1{\Gamma\left(1-\frac1\phi\right)\Gamma\left(2+\frac1\phi\right)}\tag{14}\\
&=\frac1{\Gamma\left(1-\frac1\phi\right)\Gamma\left(\frac1\phi\right)}\tag{15}\\
&=\frac1\pi\sin\left(\frac\pi\phi\right)\tag{16}
\end{align}
$$
Explanation:
$(11)$: $k^2+k-1=(k+1-\phi)(k+\phi)$
$(12)$: break up the product of factors into a product of Gamma functions
$(13)$: $\lim\limits_{n\to\infty}\frac{\Gamma(n+2-\phi)\Gamma(n+1+\phi)}{\Gamma(n+1)\Gamma(n+2)}=1$ by Gautschi's Inequality
$(14)$: $\phi=1+\frac1\phi$
$(15)$: $\Gamma\left(2+\frac1\phi\right)=\left(1+\frac1\phi\right)\frac1\phi\,\Gamma\left(\frac1\phi\right)=\phi\,\frac1\phi\,\Gamma\left(\frac1\phi\right)=\Gamma\left(\frac1\phi\right)$
$(16)$: Euler's Reflection Formula
A: Let $x_1 = t \in (0,1)$. For any $n \ge 1$, it is easy to see $x_n \ge 0$.
Since $e^y \ge 1 + y$ for all $y$, we have
$$x_n = x_1 \prod_{k=1}^{n-1} \frac{x_{k+1}}{x_k} = t \prod_{k=1}^{n-1} \left(1 + \frac{x_{k}}{k^2}\right) \le t\exp\left(\sum_{k=1}^{n-1} \frac{x_k}{k^2}\right)\tag{*1}$$
Notice


*

*$x_1 \le t$.

*if $x_n \le tn$, then $x_{n+1} \le n t + t^2 \le (n+1)t$.


By induction, we have $x_n \le n t$ for all $n$. 
Substitute this into $(*1)$, we get
$$x_n \le t \exp\left(t\sum_{k=1}^{n-1}\frac{1}{k}\right)
\le t\exp\left(t(\log n + \gamma)\right)
\le  e^\gamma n^t$$
where $\gamma$ is 
Euler-Mascheroni constant.
Substitute above into $(*1)$ again, we find
$$x_n \le t \exp\left(e^{\gamma} \sum_{k=1}^{n-1}\frac{1}{k^{2-t}}\right)
 \le t \exp\left( e^\gamma \zeta(2-t)\right) < \infty$$
where $\zeta(t)$ is the Riemann Zeta function.
A: We define sequences $(f_n)$ and $(g_n)$ of functions on the interval $(0, 1]$ such that
$$ f_{1}(x) = x, \quad f_{n+1}(x) = f_n(x) + \frac{f_n(x)^2}{n^2}, \quad g_n(x) = \frac{1}{f_n(x)}. $$
We make several observations:


*

*Since $f_n$ is positive and monotone increasing in $n$, $g_n$ is positive and monotone decreasing in $n$.

*Each $g_n$ is indefinitely differentiable on $(0, 1]$ and satisfies
\begin{align*}
g'_{n+1}(x) &= g'_n(x) \left( 1 - \frac{1}{(n^2 g_n(x) + 1)^2} \right), \\
g''_{n+1}(x) &= g''_n(x) \left( 1 - \frac{1}{(n^2g_n(x) + 1)^2} \right) + \frac{2n^2 g'_n(x)^2}{(n^2 g_n(x) + 1)^3}.
\end{align*}
Starting from $g_1'(x) = -1/x^2 < 0$ and $g_1''(x) = 2/x^3 > 0$, we can inductively prove that $g'_n \leq 0$ and $g''_n \geq 0$ for all $n$. (In other words, $g_n$ is convex and decreasing.)

*By induction, we can check that $g_n(1) = 1/n$ and $g'_n(1) = -\prod_{k=2}^{n} (1 - k^{-2})$.
Now we are ready to prove the claim. By 2, we find that for all $x \in (0, 1]$
$$ |g'_n(x)| = -g_n'(x) \geq -g_n'(1) = |g_n'(1)|. $$
Thus by the mean value theorem, there exists $c$ between $x$ and $1$ such that
$$ g_n(x) - g_n(1) = -g_n'(c)(1 - x) \geq |g'_n(1)|(1 - x). $$
Taking $n \to \infty$ and utilizing 3, we have
$$ \lim_{n\to\infty} g_n(x) \geq (1 - x) \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2} \right). $$
If we write $c = \prod_{k=2}^{\infty} (1 - k^{-2}) > 0$, this implies
$$ \lim_{n\to\infty} f_n(x) \leq \frac{1}{c(1-x)} \quad x \in [0, 1)$$
and therefore the conclusion follows.

Addendum. Let us first show that $g_n'$ converges uniformly on any compact subinterval of $(0, 1]$ as $n \to \infty$. To this end, we investigate
$$ \left| g_n'(x) - \lim_{n\to\infty} g_n'(x) \right|
= |g_n'(x)| \left[ 1 - \prod_{k=n+1}^{\infty} \left( 1 - \frac{1}{(k^2 g_k(x) + 1)^2} \right) \right]. $$
On any $[a, 1] \subset (0, 1]$, we have the following estimates:


*

*$|g_n'(x)| \leq |g_1'(x)| = x^{-2} \leq a^{-2}$.

*$g_n(x) \geq g_1(x) = 1/n$ and hence $k^2 g_k(x) + 1 \geq k+1$.
Combining these two observations, we have the following uniform estimate
$$ \left| g_n'(x) - \lim_{n\to\infty} g_n'(x) \right|
\leq \frac{1}{a^2} \left[ 1 - \prod_{k=n+1}^{\infty} \left( 1 - \frac{1}{(k+1)^2} \right) \right]. $$
It is not hard to check that this bound goes to 0 as $n\to\infty$, and hence $g_n'$ converges uniformly on $[a, 1]$. As a consequence, $g(x) = \lim_n g_n(x)$ converges uniformly on $[a, 1]$, is differentiable and  $ g'(x) = \lim_{n\to\infty} g_n'(x)$.  This shows that
$$ g'(1) = -\prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^2} \right) = -\frac{1}{2}. $$
Equivalently, we have
$$ \lim_{x \to 1^-} (1-x)f(x) = - \lim_{x \to 1^-} \frac{x - 1}{g(x) - g(1)} = -\frac1{g'(1)} = 2. $$
A: Because the sequence is increasing 
$$x_{n+1}=x_{n}+\frac{x_{n}^2}{n^2}<x_{n}+\frac{x_{n}x_{n+1}}{n^2}$$
$$\frac{1}{x_{n}}<\frac{1}{x_{n+1}}+\frac{1}{n^2}$$
(where it is easy to show that the $x_{n}>0$ is positive for all $n$)
so,
$$\sum_{i=1}^n(\frac{1}{x_{i}}-\frac{1}{x_{i+1}})<\sum_{i=1}^n\frac{1}{i^2}$$
Because of
$$\sum_{i=1}^n\frac{1}{i^2}<1+\sum_{i=2}^n\frac{1}{(i-1)i}<2-\frac{1}{n}<2$$
$$\sum_{i=1}^n(\frac{1}{x_{i}}-\frac{1}{x_{i+1}})<2$$
$$\frac{1}{x_{1}}-\frac{1}{x_{n+1}}<2$$
which is equivalent to
$$x_{n+1}<\frac{1}{\frac{1}{x_{1}}-2}$$
so the sequence is bounded above
A: Another proof. Let $x_1=x\in (0,1)$ and suppose that for all $n\ge 1$
$$x_{n+1}=x_n+{x_n^2\over n^2}.\tag{1}$$
Let's prove by induction that if  $\alpha={\ln3\over\ln2}-1\sim 0.585$ then
$$x_n<{n\over (1-x)n^\alpha+1},\qquad n\ge 1,\; x\in (0,1).\tag{2} $$
The statement is true for  $n=1$: $x_1=x<{1\over 2-x}$.
Denote  $\epsilon =1-x$ and let us prove that if  $(1)$ is true for  $n$ then it is true  for $n+1$ i.e.
$$x_n<{n\over \epsilon  n^\alpha+1}\;\;\Longrightarrow\;\; x_{n+1}<{n+1\over \epsilon (n+1)^\alpha+1},\qquad n\ge 1,\; \epsilon\in (0,1).\tag{3}$$
Using $(1)$ it's enugh to show
$${n\over \epsilon  n^\alpha+1}+{1\over n^2}\Big({n\over \epsilon  n^\alpha+1}\Big)^2<{n+1\over \epsilon (n+1)^\alpha+1}$$
that is
$$(n+1)(\epsilon  n^\alpha+1)^2-n(\epsilon (n+1)^\alpha+1)(\epsilon  n^\alpha+1)-\epsilon (n+1)^\alpha-1>0.\tag{4}$$
After a little algebra the above inequality reduces to 
$$\eqalign{&\epsilon \big(n^{\alpha+1}+2n^{\alpha}-(n+1)^{\alpha+1}\big)+\epsilon^2\big(n^{2\alpha}(n+1)-n^{\alpha+1}(n+1)^\alpha\big)\cr&=\epsilon  n^{1+\alpha}
\bigg(1+{2\over n}-\Big(1+{1\over n}\Big)^{\alpha+1}\bigg)+\epsilon^2n^{1+2\alpha}\bigg(1+{1\over n}-\Big(1+{1\over n}\Big)^\alpha\bigg).
\cr}\tag{5}$$
The coefficient of $\epsilon^2$ is clearly positive. Using the inequality $(1+t)^\beta<1+\beta t$, (true for  $t>-1$ and  $\beta\in(0,1)$) we obtain that the coefficient of  $\epsilon  n^{1+\alpha}$ satisfies
$$1+{2\over n}-\Big(1+{1\over n}\Big)^{\alpha+1}>1+{2\over n}-\Big(1+{1\over n}\Big)\Big(1+{\alpha\over n}\Big)={n-2\alpha\over n^2}$$
which is positive for  $n\ge2$.
On the other hand, when  $n=1$ the coefficient of  $\epsilon $ in $(5)$ coincides with  $3-2^{\alpha+1}$, which vanishes when $\alpha=\ln3/\ln2-1$. This proves  $(4)$ for all  $n\ge1$ and all  $\epsilon \in(0,1)$, and therefore  $(3)$ and $(2)$.
Regarding the convergence of  $x_n$, we have that for  $n\ge2$ (still with  $x=x_1$)
$$\eqalign{x_{n}&=x+\sum_{k=1}^{n-1} (x_{k+1}-x_k)<x+\sum_{k=1}^{n-1}\bigg({1\over (1-x)k^\alpha+1}\bigg)^2\cr& < x+{1\over (1-x)^2}\sum_{k=1}^\infty{1\over k^{2\alpha}}=x+{C_0\over(1-x)^2}\cr}$$
with $C_0\sim  6.4744$. The sequence $x_n$ is then bounded and increasing, so it must converge.
A: The sequense is obviously monotone increasing since $x_{n+1}>x_n$. To prove that is converging we should show that is bounded.
Expanding $x^n$ and doing the addition we take:
$$
\eqalign{
x_{n+1} &= x_n + {x_n^2 \over n^2} \cr
x_{n} &= x_{n-1} + {x_{n-1}^2 \over {(n-1)}^2} \cr
x_{n-1} &= x_{n-2} + {x_{n-2}^2 \over {(n-2)}^2} \cr
&\cdots \cr
x_2 &= x_1 + {x_1^2 \over 1^2} \cr
&\rightarrow \cr
x_{n+1} &= x_1+ {x_{1}^2 \over {1}^2}+ {x_{2}^2 \over {2}^2}+\cdots+ {x_{n-1}^2 \over {(n-1)}^2}+ {x_{n}^2 \over {n}^2} \cr
&< 1+ {2 \over {1}^2}+ {2 \over {2}^2}+ \cdots+ {2 \over {(n-1)}^2}+ {2 \over {n}^2} \cr
&<1+ {2 \pi^2 \over 6}
}
$$
A: Here's a way to show convergence without providing an explicit upper bound.
We have $$x_n=x_1+\sum_{k=1}^{n-1}(x_{k+1}-x_k)=x_1+x_1^2+\sum_{k=2}^{n-1}\frac{x_k^2}{k^2},$$ hence the convergence of the sequence would follow from that of the latter series. Now, choose $t\in\left(\frac29,1\right)$ such that $x_3<3^t$. This implies $x_2<2^t$ because otherwise we would have $2^t+2^{2t-2}<3^t$ (false for $t<1$), and inductively yields $x_k<k^{t}$ also for all larger $k$: $$\begin{align}x_{k+1}<k^{t}+k^{2t-2}&<(k+1)^{t} \\ 1+\frac1{k^{2-t}}&<\left(1+\frac1k\right)^t\\ \frac{\log(1+k^{t-2})}{\log(1+1/k)}&<t, \end{align}$$ a consequence of $$\frac{1}{k^{2-t}\log(1+1/k)}\le \frac{1}{3^{2-t}\log(3/2)}<t \iff t> \frac1{\log3} W\left( \frac{\log(3)} {9\log(3/2)} \right)\approx0.216117. $$Therefore, since $$\begin{align}\frac{x_{k+1}^2/(k+1)^2}{x_k^2/k^2}&=\left(1+\frac{x_k}{k^2}\right)^2\left(1-\frac{1}{k+1}\right)^2\\&=\left(1+\frac{2x_k}{k^2}+\frac{x_k^2}{k^4}\right)\left(1-\dfrac2{k+1}+\frac1{(k+1)^2}\right)\\&=1-\frac{2\left(1+\dfrac1k\right)^{-1}}{k}+\dfrac{\left(1-\dfrac2{k+1}\right)\left(\dfrac{x_k}{k^{1+t/2}}\right)^2+\dfrac{2x_k}{k^t}+\dfrac1{k^t}\left(1-\dfrac1{k+1}\right)^2}{k^{2-t}},\end{align}$$ the series converges by Gauss's test, ergo, so does $x_n$.
