Proof of boundedness for a sequence

Question

We are given a sequence $S_n=\sqrt{a+S_{n-1}}$ where $S_1=\sqrt a$ and $a\geq 1$.

The problem asks that we prove $S_n$ is bounded.

I am having trouble showing that it is bounded as I originally solved that $S_n = \dfrac{1+\sqrt{1+4a}}2$ as $S_n$ is essentially $\sqrt{1+S_n}$ and then solved it quadratically. Then I said that $S_n$ was bounded by $\sqrt{a+\dfrac{1+\sqrt{1+4a}}2}$

However, I think this is not right because this is equal to $S_{n+1}$ but $S_{n+2}$ is bounded by something greater than that.

Is there a way to show that this sequence is bounded?

Note: The sequence is increasing.

Answer 1

Hint: You correctly found $S_n\to(1+\sqrt{1+4a})/2$. But that is not what you were asked. You need to show $S_n < C$. It does not matter if $C$ is large. However you cannot choose $C$ to be any particular number because $a$ could be large. So how about trying to show that $S_n < 2a$? You can do this by induction.

Answer 2

Often a good way to do this is to find the limit "informally", as you have done, then prove it carefully. So, let $$L=\frac{1+\sqrt{1+4a}}{2}$$ and note that $L^2=L+a$. Then $$\eqalign{|S_n-L| &=|\sqrt{a+S_{n-1}}-L|\cr &=\Bigl|\frac{a+S_{n-1}-L^2}{\sqrt{a+S_{n-1}}+L}\Bigr|\cr &\le\frac{|S_{n-1}-L|}{\sqrt a}\cr &\le|S_{n-1}-L|\ ;\cr}$$ repeating the argument gives $$|S_n-L|\le|S_1-L|$$ and so $$L-|S_1-L|\le S_n\le L+|S_1-L|\ .$$ Since $S_1$ and $L$ are fixed numbers (in terms of $a$), this shows that the sequence is bounded.

It's not really necessary but you could write the last inequality more explicitly as $$\sqrt a\le S_n\le 1+\sqrt{1+4a}-\sqrt a\ .$$