Convergence of $a_{n+2} = \sqrt{a_n} + \sqrt{a_{n+1}}$

Question

Let $a_1$ and $a_2$ be positive numbers and suppose that the sequence {$a_n$} is defined recursively by $a_{n+2} = √a_n + √a_{n+1}$. Show that this sequence is convergent.

So, I have been able to show the convergence taking three different cases namely,

Case 1: Both $a_1$ , $a_2$ <4, then I proved that the sequence will be monotonically increasing as well as bounded and so converging

Case 2: Both $a_1$, $a_2$ >4, in this case the sequence is monotonically decreasing and bounded and hence convergent.

Case 3: One of $a_1$ and $a_2$ is <4 & other >4, lets say, $a_1$<4<$a_2$ in this case sequence will alternatively increase and decrease i.e. $a_{2n-1}$ will be increasing and $a_{2n}$ will be decreasing and $a_{2n}-a_{2n-1}$ will converge to Zero.

My question is that is there any general method through which we don't have to take all these cases and can prove the convergence of series in more generality?

Answer 1

If we can make a guess of what the limit should be, then it is often easier to show the convergence by playing with the difference between $a_n$ and the limit candidate.

In our case, the limit value must solve the constraints $x = 2\sqrt{x}$ and $x\geq 0$, hence $x = 0$ or $4$. We claim that $4$ is the limit.

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We first establish the boundedness of $(a_n)$ away from $0$ and $\infty$.

• $a_{n+2} \geq \sqrt{a_{n+1}}$ for all $n\geq1$. So $a_{n+2} \geq (a_2)^{1/2^n}$ and hence $\liminf_{n\to\infty} a_n \geq 1$.

• Let $M=\max\{4,a_1,a_2\}$. Then we inductively check that $a_n \leq M$ for all $n\geq1$.

Now define $\epsilon_n = \lvert a_n - 4 \rvert$ and $\bar{\epsilon} = \limsup_{n\to\infty} \epsilon_n$. Then $\bar{\epsilon} \in [0, \infty)$ and

$$ \epsilon_{n+2} \leq \frac{\epsilon_n}{\sqrt{a_n} + 2} + \frac{\epsilon_{n+1}}{\sqrt{a_{n+1}} + 2}. $$

Now taking $\limsup_{n\to\infty}$ to both sides yields $ \bar{\epsilon} \leq \frac{\bar{\epsilon}}{3} + \frac{\bar{\epsilon}}{3}$, which is enough to conclude that $\bar{\epsilon} = 0$ and hence $a_n \to 4$.

Answer 2

Well, if $a_n\to L$ then $L=\sqrt L+\sqrt L$, so $L=4$ or $L=0$.

So instead of trying to show the sequence converges, we try to show it converges to $4$ or $0$; that may be simpler.

Let's get rid of the square roots by defining $b_n=\sqrt{a_n}$. Now$$b_{n+2}^2=b_n+b_{n+1},$$and we want to show $b_n\to 2$. If you subtract $4$ from both sides, factor the left side and apply the triangle inequality you get $$|b_{n+2}-2|\le\frac{|b_n-2|+|b_{n+1}-2|}{b_{n+2}+2}.$$It seems likely to me you can use this to show $b_n\to2$ or $0$ (if it happens that $b_n\ge c>0$ for all $n$ then it follows that $b_n\to 2$). I have to go to class now, sorry...

Edit: How stupid of me - the other answer points out that it's more or less obvious that $b_n\ge c>0$. In case it's not clear why we care about that, it shows that $$|b_{n+2}-2|\le\frac2{2+c}\max(|b_n-2|,|b_{n+1}-2|);$$hence $b_n\to2$. since $2/(2+c)<1$.