It Telescopes?

Relevant wiki: Convergence - Ratio Test

$\begin{aligned} a_{n+1} & = \sqrt{4+3a_n+a_n^2} - 2 \\ a_{n+1} + 2 & = \sqrt{4+3a_n+a_n^2} \\ \left(a_{n+1} + 2\right)^2 & = 4+3a_n+a_n^2 \\ a_{n+1}^2 + 4a_{n+1} + 4 & = a_n^2 + 3a_n + 4 \\ a_{n+1}^2 + 4a_{n+1} & = a_n^2 + 3a_n \end{aligned}$

From the above, we note that if $a_n > 0$ , then $0 < a_{n+1} < a_n$ . Since $a_0 = 1$ , we have $0 < a_n < a_0 = 1$ . This means that $a_n$ is strictly decreasing and that as $n \to \infty$ , $a_n \to 0$ . From $a_{n+1}^2 + 4a_{n+1} = a_n^2 + 3a_n$ , and as $n \to \infty$ then $a_n^2 \to a_{n+1}^2$ , $\implies 4a_{n+1} = 3a_n$ , $\implies \dfrac {a_{n+1}}{a_n} = \dfrac 34$ .

Therefore, $\displaystyle \implies \lim_{n \to \infty} \left|\frac {a_{n+1}}{a_n} \right| = \dfrac 34 < 1$ and by ratio test $\displaystyle \sum_{n=0}^\infty a_n$ converges. Then, we have:

$\begin{aligned} a_{n+1}^2 + 4a_{n+1} & = a_n^2 + 3a_n \\ \implies a_{n+1}^2 - a_n^2+ 4\left(a_{n+1} - a_n\right) + a_n & = 0 \\ \sum_{n=0}^\infty \left(a_{n+1}^2 - a_n^2\right) + 4\sum_{n=0}^\infty \left(a_{n+1} - a_n\right) + \color{#3D99F6} \sum_{n=0}^\infty a_n & = 0 \\ - a_0^2 - 4a_0 + \color{#3D99F6} S & = 0 \\ \implies S & = {\color{#3D99F6}a_0}^2 + 4{\color{#3D99F6}a_0} & \small \color{#3D99F6} \text{Note that } a_0 = 1 \\ & = 5 \end{aligned}$

$\implies \lfloor 100S \rfloor = \boxed{500}$

Relevant wiki: Telescoping Series - Sum

In the form given by the problem, we can see that if $a_{n}>0$ , then $a_{n+1}>\sqrt{4}-2=0 \Rightarrow a_{n+1}>0$ . Since $a_{0}>0$ , it follows that $a_{n}>0$ by induction on $n$ .

Now, rearrange the recurrence relation to get $\left( a_{n+1}+2 \right)^{2}= \left( a_{n}+2 \right)^{2} - a_{n}$ . In this form, it is easy to see that $(a_{n+1}+2)^{2}<(a_{n}+2)^{2}$ since $a_{n}>0$ . It follows that $a_{n+1}<a_{n}$

We now know that $a_{n}$ is a strictly decreasing sequence of positive reals, and so $a_{n}$ must converge to some limit $0 \leq L < a_{0}$ . Thus, we have that $(L+2)^{2}=(L+2)^{2}-L \Rightarrow L=0$

Rearrange again to get $(a_{n}+2)^{2}-(a_{n+1}+2)^{2}=a_{n}$ , and so $S$ can be written as a telescoping sum:

$S \ = \ \sum_{n=0}^{\infty} a_{n} \ = \ \sum_{n=0}^{\infty} [(a_{n}+2)^{2}-(a_{n+1}+2)^{2}] \ = \ \lim_{p \rightarrow \infty} (a_{0}+2)^{2} - (a_{p}+2)^{2} \ = \ 5$

So $\lfloor 100S \rfloor = \boxed{500}$

Relevant wiki: Convergence - Ratio Test

While the other solutions have arrived at the correct limit, I don't believe the mechanics behind either are entirely rigorous.

First, it can be easily shown through induction that $a_n>0, \forall n \in \mathbb{N}$ .

Then, $\displaystyle{\sqrt{4+3a_n+{a_n}^2}<\sqrt{4+4a_n+{a_n}^2}\implies \frac{a_{n+1}}{a_n}<1\implies a_{n+1}<a_n}$ .

Since $a_n$ is decreasing and bounded below, then there is some $a\geq 0$ such that $a_n \rightarrow a$ as $n \rightarrow \infty$ .

Now we can treat this limit algebraically. From $\displaystyle{a_{n+1} = \sqrt{4+3a_n+{a_n}^2}-2}$ , send $n$ to infinity on both sides and we get

$\displaystyle{a = \sqrt{4+3a+{a}^2}-2 \implies 4a=3a \implies a = 0}$ .

To see if $\displaystyle{\sum a_n}$ converges, we can perform the ratio test:

$\displaystyle{\left|\frac{a_{n+1}}{a_n} \right| = \left|\frac{\sqrt{4+3a_n+{a_n}^2}-2}{a_n} \cdot \frac{\sqrt{4+3a_n+{a_n}^2}+2}{\sqrt{4+3a_n+{a_n}^2}+2}\right| = \left|\frac{3+a_n}{\sqrt{4+3a_n+{a_n}^2}+2} \right| }$ , since $a_n \neq 0$ .

Since $a_n \rightarrow 0$ , then $\displaystyle{\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n} \right| = \left| \frac{3+0}{\sqrt{4+3(0)+0^2}+2} \right| = \frac{3}{4}}$ .

Since this limit is strictly less than $1$ , then $\displaystyle{\sum a_n}$ converges absolutely to some real number. We can now treat THAT limit algebraically as well.

Let $\displaystyle{A = \sum a_n}$ . Then $\displaystyle{A -1 = \sum a_{n+1}}$ .

Since $\displaystyle{\sum a_n}$ converges and all $a_n > 0$ , then there must be some $N$ such that for all $n\geq N$ we have $a_n <1$ .

This means that for all $n\geq N$ we have $\displaystyle{{a_n}^2<a_n}$ ; i.e. $\displaystyle{\sum {a_n}^2}$ converges as well.

Let $\displaystyle{B = \sum {a_n}^2}$ . Then $\displaystyle{B -1 = \sum {a_{n+1}}^2}$ .

Finally, we have

$\displaystyle{a_{n+1} = \sqrt{4+3a_n+{a_n}^2}-2 \implies {a_{n+1}}^2+4a_{n+1}= {a_{n}}^2+3a_{n} \implies \sum {a_{n+1}}^2+4\sum a_{n+1}= \sum {a_{n}}^2+3\sum a_{n} \implies B-1+4(A-1)= B+3A \implies A=5}$