Sequence limit and monotony of $a_{n+1}=\sqrt{4a_n+3},a_1=5$ I am having difficulty proving whether the sequence is increasing or decreasing and finding its limit. Using derivative I find that it is increasing, but there seems to be another way. Computing $a_{n+1}-a_n$ gets messy for me.
$$a_{n+1}=\sqrt{4a_n+3},a_1=5$$
 A: You don't really need to prove that the series is monotonic.
If you assume that the series converges you can use a quite common trick to find the limit. Say the limit exists and it's equal to $L$
$$\lim_{n\to\infty} a_n=L$$
Intuitively you know that the limit for $a_n$ and for $a_{n+1}$ should be the same so you can write something like
$$\lim_{n\to\infty} a_n=L=\lim_{n\to\infty} a_{n+1}=\lim_{n\to\infty} \sqrt{4a_n+3}$$
Now, as the limit for $a_n$ exists you can use some property of limits of functions to get to
$$\lim_{n\to\infty} \sqrt{4a_n+3}= \sqrt{4\lim_{n\to\infty}a_n+3}=\sqrt{4L+3}$$
So you end up with
$$L=\sqrt{4L+2}$$
That gives you two solutions for $L$, $2+\sqrt{7}$ and $2-\sqrt{7}$. Now, as you know the series is all positives (because square roots are always positive numbers) then you can discard the negative solution and get
$$L=2+\sqrt{7}$$
A: We can prove by induction to show that the sequence $\{a_n \}$ is decreasing.
For $n=1$, we have $$a_2=\sqrt{4a_1+3}=\sqrt{4(5)+3}=\sqrt{23}<\sqrt{25}=5=a_1,$$ or $a_1>a_2.$ 
For $n=k$, assume that the statement $$a_k>a_{k+1} \tag{$\star$}$$ is true. Now, multiplying both sides by $4$, then adding $3$ to both sides, then applying the square root to both sides of $(\star)$ gives $$\sqrt{4a_k+3}>\sqrt{4a_{k+1}+3}.$$ This means $$a_{k+1}>a_{k+2}.$$
and hence the sequence is decreasing.

Now, to find the limit, set $a=\lim_{n\to \infty} a_n$. Then $a=\lim_{n\to \infty} a_{n+1}$ as well. Thus,
\begin{align}
\lim_{n \to \infty} a_{n+1}=\lim_{n\to \infty} \sqrt{4a_n+3}
\end{align}
becomes
\begin{align}
a= \sqrt{4a+3}.
\end{align}
Now solve for $a$ and you will find your limit.
A: the sequence $\{a_n \}$ is decreasing and converges to $2 + \sqrt 7$ 
we also have $$(a_{n+1} - 2 - \sqrt 7) = 
\dfrac{4(a_n - 2 - \sqrt 7)}{\sqrt{4a_n + 3} + 2 + \sqrt 7}<\dfrac{4}{2+\sqrt 7}(a_n-2-\sqrt 7)$$ which guarantees the convergence and monotonicity of $a_n$.
A: If $a_{n-1}<a_n$ then $\sqrt{4a_{n-1}+3}<\sqrt{4a_n+3}$, so $a_n<a_{n+1}$.
