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?


marked as duplicate by Gabriel Romon, Martin R, Sil, Siméon, Sangchul Lee real-analysis May 2 '18 at 18:54

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    $\begingroup$ The things you say you proved under Case 1 and Case 2 cannot both be true. (I don't see why any of the three is true, actually...) $\endgroup$ – David C. Ullrich May 2 '18 at 15:35
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    $\begingroup$ @DavidC.Ullrich: He is splitting up the question into cases. In the first possible case, we have $a_1,a_2<1$. In a second possible case, we have $a_1,a_2>1$. In a third possible case, we have $a_1<1<a_2$ or vice versa. No matter the sequence $a_i$, $a_1$ and $a_2$ must be in one of these three cases. $\endgroup$ – Clayton May 2 '18 at 15:39
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    $\begingroup$ If $a_n$ converges to $l$ then $l=2 \sqrt{l}$. So $l=0$ or $l=4$. I think this is why @DavidC.Ullrich say that both affirmation cannot be true, if Case $1$ is true then for $n$ large enough $a_n>1$ which is problematic with Case 2. $\endgroup$ – Delta-u May 2 '18 at 15:54
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    $\begingroup$ @Clayton I understood he's splitting it into cases... that doesn't explain why his conclusions in those cases are correct. $\endgroup$ – David C. Ullrich May 2 '18 at 16:05
  • $\begingroup$ Sorry, I made a mistake. I should have written 4 instead of 1. Edited it. $\endgroup$ – Lord KK May 2 '18 at 16:07

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.

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 $a_n \to 4$.


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 $2/(2+c)<1$.

  • $\begingroup$ But how can you claim the boundedness on the right hand side? $\endgroup$ – Lord KK May 2 '18 at 16:13
  • $\begingroup$ @AloknathFurr I'm not sure what you're referring to. If $b_n\ge c>0$ then the inequality shows that $|b_n-2|$ decreases geometrically. $\endgroup$ – David C. Ullrich May 2 '18 at 16:15
  • $\begingroup$ ok.. ok.. I got it. Thanks $\endgroup$ – Lord KK May 2 '18 at 16:17

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