Let $a>0$ and $x_1 > 0$ and $x_{n+1} = \sqrt{a + x_n}$ for $n \in \mathbb{N}$. Show that $\{x_n\}_{n\ge 1}$ converges

Question

Let $a>0$ and $x_1 > 0$ and $x_{n+1} = \sqrt{a+x_n}$ for $n \in \mathbb{N}$. Show that $\{x_n\}_{n\ge 1}$ converges

Please help me with this problem. I am unable to prove it monotone and bounded.

Answer 1

Let $L=\frac{1}{2}(1+\sqrt{4a+1}), L'=\frac{1}{2}(1-\sqrt{4a+1})$. Clearly $x_n>0$. Suppose $x_1\le L$. Then if $x_n\le L$, one has $$ x_{n+1}=\sqrt{a+x _n}\le\sqrt{a+L}=L. $$ So $\{x_n\}$ is bounded. Note $$ x_{n+1}-x_n=\sqrt{a+x_n}-x_n=\frac{a+x_n-x_n^2}{\sqrt{a+x_n}+x_n}=-\frac{(x_n-L)(x_n-L')}{\sqrt{a+x_n}+x_n}\ge 0$$ which shows that $\{x_n\}$ is increasing. So $\{x_n\}$ is bounded and increasing and hence there is $x$ such that $\lim_{n\to\infty}x_n=x$. Letting $n$ go to infinity in $x_{n+1}=\sqrt{a+x_n}$, one has $$ x=\sqrt{a+x} $$ which implies $x=L$.

One can do the same thing for $x_1> L$.

Answer 2

Have you tried this?

$$\lim_{n\to\infty}\dfrac{T_{n+1}}{T_n}=0$$

Answer 3

Let $x=1/2+\sqrt{1/4+a}$. Then, if $y<x$, it can be readily verified that $y^2-y=(y-1/2)^2-1/4<a$, and $y^2-y>a$ if $y>x$. Hence, if $x_n<x$, $x_{n+1}=\sqrt{x_n+a}>x_n$, and $x_n\ge x\implies x_{n+1}\le x$. Thus, if $x_1<x$, the sequence $x_n$ is a positive increasing sequence upper bounded by $x$, and if $x_1\ge x$, $x_n$ is a positive decreasing sequence lower bounded by $x$. Hence the sequence converges in either of these cases, and it converges to $x$.

Answer 4

D'Alembert's criterion

Calculate limit ratio of next to preceding elements.

$\lim_{n\rightarrow\infty}{\frac{x_{n+1}}{x_{n}}}=\lim_{n\rightarrow\infty}{\frac{\sqrt{a+x_{n}}}{x_{n}}}=r$

If $|r| < 1$ , then the series converges.

Hints for solving the limit: extend by the square root, divide by $x_n$, separate into nominator and denominator limits, solve separately, resulting in $\frac{1}{\infty}$.

Answer 5

Yet another (partial) approach:

Consider $f(x) = \sqrt{a + x} \ \forall x > 0$ for some $a > 0$. Then $f'(x) = \frac 1 {2\sqrt{a + x}}$, so

$$ x > 0.25 - a \implies \sqrt{a + x} > 0.5 \implies \frac 1 {2\sqrt{a + x}} = f'(x) < 1 $$

, so $f$ is contractive in $]0.25 - a, + \infty[$. If $x_0 > 0.25 - a$ then the sequence converges.

If $a \ge 0.25$ any $x_0 > 0 \ge 0.25 - a$ will converge in that fashion. Otherwise, if $x_0 \in ]0, 0.25 - a[$ (and so $a < 0.25$), then $$ f(x) = \sqrt{a + x} > \sqrt{x} > x \ \forall x \in \ ]0, 0.25 - a[ \ \subset \ ]0, 0.25[ $$

so the sequence is strictly increasing and will also converge.

Answer 6

If there is a limit, it must be a fixed point:

$$\begin{eqnarray} y && = && \sqrt{a+y} \\ y^2 && = && a+y \\ y^2 - y - a && = 0 \\ y && = && \frac{1 \pm \sqrt{1+4a}}{2} \end{eqnarray}$$

We want the positive answer, obviously.

Next show the fixed point is attractive. Let $e_n = x_n - y$. Then:

$$\begin{eqnarray} e_{n+1} && = && \sqrt{a+y+e_n} - y\\ && = && \sqrt{a+y}\sqrt{1+\frac{e_n}{a+y}} - y \\ && = && y\left(1 + \frac{e_n}{2(a+y)}+ O(e_n^2) \right) - y \\ && = && \lambda e_n + O(e_n^2) \end{eqnarray}$$

where $\lambda = \frac{y}{2(a+y)}$. It is attractive provided $|\lambda|<1$, which is always true because $a$ and $y$ are positive.

Finally, we need to show convergence for any $x_1$, not just for values near $a$. This follows from the fact that $\sqrt{a+x}$ is a continuous function. If there was an $x_1$ that diverges, then there would be a value between there and $a$ that is another fixed point, but we have already shown that there is none.