Prove that the recursive sequence $(a_n)_{n \ge 1}$ is convergent and find its limit I have the sequence $(a_n)_{n \ge 1}$ defined as $a_{n+1}=\sqrt{12+4a_n}$ with $a_1=1$ and I'm trying to:


*

*prove that it converges &

*find its limit.


My attempt:


*

*First, I assume that $(a_n)_{n \ge 1}$ is increasing and bounded above, thus being convergent and:
$$
a_1 \le a_k \le \lim_{n→\infty}a_n
\Leftrightarrow
1 \le a_k \le L
$$

*Then, I calculate $L$ using the definition of $(a_n)_{n \ge 1}$:
$$
L=\sqrt{12+4L}
\Leftrightarrow
L^2-4L-12=0
\Leftrightarrow
(L+2)(L-6)=0
\Leftrightarrow
L \in\{\require{cancel} \cancel{-2}, 6\}
$$

*Finally, I prove that the initial assumption is correct:
$$
a_{n+1} \ge a_n
\Leftrightarrow
\sqrt{12+4a_n} \ge a_n
\Leftrightarrow
a_n^2-4a_n-12 \le 0
\Leftrightarrow\\
(a_n+2)(a_n-6) \le 0
\Leftrightarrow
-2 \le a_n \le 6
$$
So, in order for $(a_n)_{n \ge 1}$ to be increasing $a_n \le 6 \Leftrightarrow a_n \le L$, which is true.


Questions:


*

*Is my solution adequate at proving that $(a_n)_{n \ge 1}$ converges and at calculating its limit?

*Is there a better way to solve this problem?

 A: Such problems are easily solved by drawing a picture. Plot the graph of the functions $y = \sqrt{12 + 4x}$ and $y = x$. 

Your sequence will be a sequence of such points (I'm sorry, I'm not good at painting)
I think you will be able to guess from that picture which point this sequence converges to
A: Criticisms: 
$A=B$ is not equivalent to $A^2=B^2$ in general, but it is when it is known that  $A,B$ are not negative... Since $\sqrt {\cdot}\;$ is by definition, never negative, you should write $$L=\sqrt {12+4L}\; \iff (L\geq 0\land L^2=12+4L)\iff $$ $$\iff (L\geq 0\land (L+2)(L-6)=0)\iff$$ $$\iff (L\geq 0\land L\in \{-2,6\})\iff L=6.$$
You should not assume the sequence is bounded above. You can prove it, preferably by induction: (1). $0<a_1<6.$.... (2). If $0<a_n<6$ then $0<\sqrt {12}\;<\sqrt {12+4a_n}\;<\sqrt {12 +(4)(6)}\;=6.$....   BTW this also shows that the sequence $(a_n)_n$  of real numbers actually exists. (I.e. if $a_1\geq 0$ then $\forall n\;(12+4a_n\geq 0).$
As in my 1st paragraph, the assertion that $a_{n+1}\geq a_n$ is equivalent to $a_{n+1}^2\geq a_n^2$ is valid only if you know that  neither $a_{n+1}$ nor $a_n$ is negative.  This is known from part of the proof in my 2nd paragraph above.  
Praises:
You have exactly the right ideas. One way to improve your expositional skills is to (sometimes tediously) examine each and every "implies " and "iff "  and ask  yourself how it is justifiable. (Your skills are already good. Many  students, unlike you, write in disconnected sentences without any $\implies$ or $\iff.$).... And never state an unproven assumption. At a university level, markers of assignments will subtract big marks.
I don't think there is an easier method. A descriptive summary would be : By  the recursive formula, if $L=\lim_{n\to \infty}a_n$ exists then $L=6.$ And by induction on $n,$ if  $0\leq a_1\leq 6$ then $\forall n \;(0\leq a_n\leq a_{n+1}\leq 6.$  So if $a_n=1$ then $\lim_{n\to \infty}a_n$ exists.  Therefore  $L=6.$ 
A: My understanding (if this is for an intro to analysis / advanced calculus course) is that this is the intended approach to solving the problem.
I would say that the only part that you might want to fix up is how you conclude that $a_n \leq 6$ for all $n$. Remember, we know that is converges to $6$ via monotonically increasing only if $a_n$ is monotonically increasing (this is step 1). However, if it happened that $a_{30} \leq 6$ but then $a_{30} < a_{31} = 7700$, we would have a problem. I would include along with part 3 a proof by induction that $a_n \leq 6$ for all $n$. In particular, note that in part 3 you never proved that $a_n$ was bounded (at all), which was necessary for step 1.
Another proof that one might see (perhaps later in the course after discussing Cauchy sequences and complete metric spaces) is via using the contraction mapping principle, which basically easily solves all sorts of problems like the one that you are looking at and also other ones that are trickier to work out "with your bare hands".
A: The graphical approach is pretty useful here. From a higher point of view, we may prove that the function $f(x)=\sqrt{12+4x}$ is a contraction of the metric space $I=[0,7]$. Indeed
$$ f'(x) = \frac{1}{\sqrt{3+x}}\leq \frac{1}{\sqrt{3}} < 1$$
implies that $f(x)$ has a unique fixed point in $I$ and the sequence $f(1),f(f(1)),f(f(f(1))),\ldots$ is convergent to such fixed point. The only solution of $f(x)=x$ in $I$ is $x=6$, hence $\lim_{n\to +\infty}a_n=6$.
This approach allows you not to discuss the monotonicity of $\{a_n\}_{n\geq 1}$.
A: In order to prove $a_n$ converges, all we have to do is prove it's bounded above and increasing:


*

*First, we'll need to prove $a_n$ is bounded above by $6$ through induction. For $n=1$ it holds, so let's assume it holds for $k=n\ge 1$ as well:
$$
a_k<6\Leftrightarrow
4a_k<24\Leftrightarrow
12+4a_k<36\Leftrightarrow
\sqrt{12+4a_k}<\sqrt{36}\Leftrightarrow
a_{k+1}<6
$$

*Then, we'll need to prove that  $a_n$ is increasing through induction. For $n=2$ it holds, because $a_1<a_2=1<\sqrt{16}=4$, so let's assume it holds for $k=n\ge 1$ as well:
$$
a_k<a_{k+1}\Leftrightarrow
12+4a_k<12+4a_{k+1}\Leftrightarrow
\sqrt{12+4a_k}<\sqrt{12+4a_{k+1}}\Leftrightarrow
a_{k+1}<a_{k+2}
$$
Since  $a_n$ is bounded above and increasing, it's bounded, convergent and $a_1<a_k<\lim_{n\to\infty}a_n$. The limit can be found easily by substituting it in the place of $a_i$:
$$
L=\sqrt{12+4L}\Leftrightarrow
L\ge 0~~\land~~L^2-4L-12=0\Leftrightarrow
L\ge 0~~\land~~(L+2)(L-6)=0
$$
$$
\Leftrightarrow
L\ge 0~~\land~~L \in \{-2, 6\}\Leftrightarrow
L=6
$$
