if $S_{n+1}=\frac{4+3S_{n}}{3+2S_{n}}$ find $\lim_{n \to \infty} S_{n}$

Question

A sequence $\left\{t_n\right\}$ is defined as $t_1=1$ and $$S_{n+1}=\frac{4+3S_{n}}{3+2S_{n}}$$ where $S_{n}=\sum t_n$

find $$\lim_{n \to \infty} S_{n}$$

I have found $(n+1)$th term as

$$t_{n+1}=S_{n+1}-S_n=\frac{4-2S_n^2}{3+2S_n}$$ From this we have

$t_1=1$, $t_2=\frac{2}{5}$, $t_3=\frac{2}{145}$ and so on. But i could not find any pattern in the sequence

Answer 1

The limit approaches $\sqrt{2}$. A common method for calculating this is to set $S_{n+1}=S_{n} = S$, and finding the solution to $$S = \frac{4+3S}{3+2S}$$ which is $S=\sqrt{2}$.

Note: Let's add a proof that $S_n$ is increasing and bounded above and hence converges. Notice that when $1\leq S_n\leq \sqrt{2}$, $S_{n+1} = \frac{4+3S_n}{3+2S_n}>S_n$ (equivalent to $S_n^2<2)$. Hence, $S_n$ is increasing. Further, let's prove that this sequence is always bounded above by $\sqrt{2}$. This is proven by the fact that when $S_n\leq \sqrt{2}$, $$S_{n+1} = \frac{4+3S_n}{3+2S_n} = \frac{3}{2}-\frac{1}{6+4S_n} \leq \frac{3}{2}-\frac{1}{6+4\sqrt{2}} = \sqrt{2}$$ Hence, $S_n$ converges and our method is correct.

Answer 2

Let $f(x) = \frac{4+3x}{3+2x}$, then $$f^\prime(x) = \frac{1}{(3+2x)^2}>0.$$

So $f$ is increasing. Now, $$S_1 = 1, S_2=f(S_1) = 7/5>S_1$$ so $S_n$ is an increasing function. So if $\{S_n\}$ is bounded above, it is convergent.

All you have to do is to find the smallest solution $\alpha>1$ for $$x=f(x) = \frac{4+3x}{3+2x}.$$ This is because $\alpha>S_n$ for every $n$, so $S_n$ converges to $\alpha$

Answer 3

$S_1=1$, which gives $S_n>0$ and since $S_{n+1}-\sqrt2=\frac{4+3S_n}{3+2S_n}-\sqrt2=\frac{(3-2\sqrt2)(S_n-\sqrt2)}{3+2S_n}$,

by induction we obtain that $S_n<\sqrt2$ for all $n\geq1$.

In another hand, $$S_{n+1}-S_n=\frac{4+3S_n}{3+2S_n}-S_n=\frac{2(\sqrt2+S_n)(\sqrt2-S_n)}{3+2S_n}>0,$$ which says that $S$ is an increasing.

Thus, there is $\lim\limits_{n\rightarrow+\infty}S_n$ and let this limit be $a$.

Thus, $a=\frac{4+3a}{3+2a}$ and since $a>0$, we obtain $a=\sqrt2$.