Show that the following inductively defined sequence converges and find its limit.

Question

Let a > 0 and let $x_1 > 0$. Define $x_{n+1} := \sqrt{a + x_n}$ for n $\in$ N. Show that <$x_n$> converges and find its limit.

Answer 1

$$x_{n+1}-x_n=\sqrt{a+x_n}-x_n=\frac{a+x_n-x_n^2}{\sqrt{a+x_n}+x_n}=$$ $$=\frac{\left(\frac{1+\sqrt{1+4a}}{2}-x_n\right)\left(\frac{\sqrt{1+4a}-1}{2}+x_n\right)}{\sqrt{a+x_n}+x_n}.$$ Now, by induction we see that if $x_1>\frac{1+\sqrt{1+4a}}{2}$ then $\{x_n\}$ is decreasing

and $x_n>\frac{1+\sqrt{1+4a}}{2}$ for all $n$.

Thus, there is $\lim\limits_{n\rightarrow+\infty}x_n$.

If $x_1<\frac{1+\sqrt{1+4a}}{2}$ then $\{x_n\}$ is increasing

and $x_n<\frac{1+\sqrt{1+4a}}{2}$ for all $n$.

Thus, there is $\lim\limits_{n\rightarrow+\infty}x_n$.

Let $\lim\limits_{n\rightarrow+\infty}x_n=x$.

Thus, $x=\sqrt{a+x}$ and $x=\frac{1+\sqrt{1+4a}}{2}$.

Done!

Answer 2

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.

Answer 3

Hint: Since both $a>0$ and $x_{1}>0$ we know that $x_{n}>0$ as well. Additionally, if the limit exists, we can say that

$$\lim_{n\to\infty}x_{n+1} = \sqrt{a + \lim_{n\to\infty}x_{n}}$$

Denoting $\lim_{n\to\infty}x_{n+1} = x = \lim_{n\to\infty}x_{n}$ we see that

$$ x = \sqrt{a+x} \implies x^{2} - x - a = 0 \implies x = \frac{1 \pm\sqrt{1+4a}}{2}, $$

and as $x\ge 0$ we know that $$x=\frac{1+\sqrt{1+4a}}{2}.$$

Now you just need to argue that the limit does indeed exist.