Convergence of $u_{n} = \frac{1}{n}+\sqrt{u_{n-1}}$ with $u_0\geqslant0$

Question

I'm trying to know for which $u_0\geqslant 0$ the following sequence $$ \forall n\in \mathbb{N},\qquad u_{n} = \frac{1}{n}+\sqrt{u_{n-1}} $$ converges. One can see that : if $u_0\leqslant v_0$ then $u_n\leqslant v_n$ for all n ; the only possible limit is $1$ (since $u_n \geqslant 1$ for all $n>0$) ; if $u_0\leqslant 3$, $\lim u_n =1$ (since $u_n \leqslant 1+3/n$ for $n>1$ when $u_0=3$).

Numerically, it seems to converge for every $u_0$ to $1$ but how can I find a proof for large $u_0$ ?

Answer 1

From $u_n\ge 1$ for each $n\ge 1$, we deduce that $u_n\le 2\sqrt{u_{n-1}}$. This means $$ {u_1\le 2\sqrt{u_0} \\ u_2\le 2\sqrt{u_1}\le 2\sqrt{2\sqrt{u_0}}=2^{1+\frac{1}{2}}u_0^\frac{1}{4} \\ u_3\le 2\sqrt{u_2}\le\cdots\le2^{1+\frac{1}{2}+\frac{1}{4}}u_0^\frac{1}{8} \\\vdots } $$ In general, $u_n\le 2^{2-2^{1-n}}u_0^\frac{1}{2^n}\le 4u_0^\frac{1}{2^n}$. Since $\lim_{n\to\infty}u_0^\frac{1}{2^n}=1$, for any initial value $u_0$, there is some $n_0$ such that $u_{n_0}\le 9$ ($9$ is arbitrary; any other value from $\Bbb R^{>4}$ is acceptable). Therefore, $$ u_{n_0}\le 9 {\implies u_{n_0+1}\le1+\sqrt{u_{n_0}}=4 \\\implies u_{n_0+2}\le1+\sqrt{u_{n_0+1}}=3 \\\implies u_{n_0+3}\le1+\sqrt{u_{n_0+2}}=1+\sqrt{3}<3. } $$ Now, the same proof of convergence for $u_0< 3$ can be conducted here as well, since we have proved that the sequence $u_n$ finally reduces below $3$.

Answer 2

It's easy to get: $u_n\geq1$ for each $n\geq1$.
First, if $\{u_n\}$ is convergent, let $x$ be its limit, then $x=\sqrt{x},\ x\geq1$, and this implies $x=1$.
Second, we have the following key inequality: $$u_{n+1}-u_n=\frac{-1}{n(n+1)}+\sqrt{u_n}-\sqrt{u_{n-1}} <\frac{u_n-u_{n-1}}{\sqrt{u_n}+\sqrt{u_{n-1}}},\quad n\geq1.$$ Due to this key inequality, we know that: for any given $u_0\geq0$, there exists $N\in\mathbb{N}$, such that $u_{_{N+1}}\leq u_{_N}$. Otherwise, the sequence $\{u_n\}_{n\geq0}$ is strictly increasing which is implossible:

If $\{u_n\}_{n\geq0}$ is strictly increasing:
Case one: If $\{u_n\}_{n\geq0}$ is unbounded above, then, when $n$ large enough, $$u_n=\frac{1}{n}+\sqrt{u_{n-1}}<u_{n-1}.$$ This is a contradiction.
Case two: If $\{u_n\}_{n\geq0}$ is bounded above, then $\{u_n\}$ is convergent and the limit is $x=1$, but $x\geq u_2=\frac{1}{2}+\sqrt{u_1}\geq\frac{3}{2}$, also a contradiction.

So, for any given $u_0\geq0$, there exists $N\in\mathbb{N}$, such that $x_{_{N+1}}\leq x_{_N}$. This implies the sequence $\{u_n\}_{n\geq N}$ is deceasing. Conbining with $u_n\geq0$, we know $\{u_n\}$ is convergent.