Shoulders, Giants, and Cats🐈

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

In the following problems, we find the limits of sequences $\{a_n\}$ given the functional form of the sequences. We assume that some fundamental limits such as $$ \lim_{n\rightarrow\infty}\frac{1}{n} = 0 $$ and $$ \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n \equiv e $$ are known. $a_n = \left(1-\frac{1}{n}\right)^n $ $$\begin{align} a_n &= \left(\frac{n-1}{n}\right)^n% = \left(\frac{n}{n-1}\right)^{-n}\\ &=\left(\frac{n-1}{n-1}+\frac{1}{n-1}\right)^{-n}\\ &=\left(1+\frac{1}{n-1}\right)^{-n}\\ &=\left(1+\frac{1}{n-1}\right)^{-(n-1)}\left(1+\frac{1}{n-1}\right)^{-1}\\ &=\frac{\left(1+\frac{1}{n-1}\right)^{-1}}{\left(1+\frac{1}{n-1}\right)^{(n-1)}}\\ % \lim_{n\rightarrow\infty}a_n% &=\lim_{n\rightarrow\infty}\frac{\left(1+\frac{1}{n-1}\right)^{-1}}{\left(1+\frac{1}{n-1}\right)^{(n-1)}}\\ &=\frac{\lim_{n\rightarrow\infty}\left(1+\frac{1}{n-1}\right)^{-1}}{\lim_{{n-1}\rightarrow\infty}\left(1+\fra