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Find the limit of the sequence. $a_1=1, a_{n+1} = \frac{3a_n+4}{2a_n+3}$

We take the limits of sequences defined recursively by first showing the existence of the limits then, actually computing them. Existence is demonstrated by showing that the sequence is both (1) monotonic and (2) bounded.

Problem

Find the limit of the following sequence. $$\begin{align} $a_1&=1\\ a_{n+1} &= \frac{3a_n+4}{2a_n+3}$ \end{align}$$

Solution

  • Monotonicity: We use mathematical induction to show that for all $n\in\mathbb{N}$, $a_{n+1} - a_n \geq 0$ and therefore, $\{a_n\}$ is monotonic increasing.

    For the base case at $n=1$, it is easy to see that $a_{n+1}-a_n = \frac{7}{5} - 1 = \frac{2}{5} >0.$

    For the inductive case at $n=k$, we assume that $$\begin{align} a_{k+1} - a_{k} & = \frac{3a_k+4}{2a_k+3} - \frac{3a_{k-1}+4}{2a_{k-1}+3}\\ & = \frac{a_k - a_{k-1}}{(2a_k+3)(2a_{k-1}+3)} \geq 0. \end{align}$$ For $n=k+1$, we get, $$\begin{align} a_{k+2} - a_{k+1} & = \frac{a_{k+1} - a_{k}}{(2a_{k+1}+3)(2a_{k}+3)} \\ &\geq 0\\ &\because a_{k+1} - a_{k} >0.□ \end{align}$$

  • Boundedness: Because the sequence is monotonic increasing, and $a_1=1$, then $a_n \geq 1$. Therefore, the sequence is bounded below.

    To show that the sequence is also bounded above, we observe that $$\begin{align} a_{n} &= \frac{3a_{n-1}+4}{2a_{n-1}+3} \\ &\leq\frac{3a_{n-1}+4}{a_{n-1}} \because{\forall n\in\mathbb{N}, a_n\geq1}\\ &\leq 3 + \frac{4}{a_{n-1}} \leq 7.□ \end{align}$$

  • Limit: If the limit $L \equiv \lim_{n\rightarrow\infty} a_n$ exists, then $$\begin{align} \lim_{n\rightarrow\infty}a_{n+1} &= \lim_{n\rightarrow\infty}\frac{3a_n+4}{2a_n+3}\\ L &= \frac{3L+4}{2L+3}\\ 2L^2 + 3L &= 3L +4\\ L &= \pm\sqrt{2} \\ \Rightarrow L &=\sqrt{2} \because L>0.■ \end{align}$$

References

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