How to prove that the limit of sequence $a_n$ exist?

Question

Given $a_1 =1$, $a_2=2$ and $$a_{n+1}=\sqrt{a_n}+\sqrt{a_{n-1}}.$$

To find the limit, I assumed that $a_n =a_{n+1}=a_{n-1}$, as $n$ tends to $\infty$ but this can be assumed if the sequence converges. How to prove that this sequence is convergent?

Answer 1

At first glance, the sequence looks like it might be increasing. If that is the case, to be convergent it would need to be bounded (and conversely if it is bounded and increasing then it must converge). The recursive definition lends itself naturally to induction. I.e., we want to prove $a_{n+1}\geq a_n$ for all $n\geq1$.

For the base case, $a_2=2\geq1=a_1$. For the inductive bit, you can pretty quickly find $a_{n+2}\geq2\sqrt{a_n}$, but we really want this to be bounded below by $a_{n+1}$ to finish the induction. It isn't immediately obvious why that should be the case, so I'm just going to amend the induction we're doing to include the extra hypothesis and start over.

Instead of just proving that the sequence is increasing, we're going to prove the bound $$\left(\frac{a_{n+1}}{2}\right)^2\leq a_n\leq a_{n+1}.$$ Note that it's trivially true for the base case $n=1$, and as luck would have it an inductive proof is straightforward (proof left to the reader).

Since the inequality is true for all $n\geq1$, we just need to prove that the sequence is bounded to prove that it converges. Ignoring the middle part of the inequality, we already know $$\left(\frac{a_{n+1}}{2}\right)^2\leq a_{n+1}.$$ Solving that inequality we find $0\leq a_{n+1}\leq4$. Hence, our sequence is convergent.