Consider the inductive sequence $(y_n)$ defined by $$y_1= 1\hspace{20pt} y_{n+1}= \dfrac{1}{4}(2y_n+3)\hspace{4pt}\forall n\in\mathbb{N}$$ The author first proved by induction that it is bounded and monotonically increasing thus its convergent, and then he says although the limit of a bounded monotone sequence can be calculated by evaluating $\sup\{y_n: n\in\mathbb{N}\}$ but since it would be a complicated process here he took the help of algebraic limit theorems.
But what if I treat it as a direct question, like find $\sup \{y_n: n\in\mathbb{N}\}$ where $y_n$ is defiend as above. Clearly we have a upper bound $2$ of the set, but I'm not sure what approach should I take to go for the least upper bound.
As $y_{n+1} -\frac{3}{2} = \frac{1}{2} \left( y_{n} -\frac{3}{2} \right)$, you can deduce the closed form expression of $y_n$ as $$y_n = \frac{3}{2}+\frac{1}{2^{n-1}}\left( y_{1} -\frac{3}{2} \right) =\frac{3}{2}-\frac{1}{2^{n}}$$
And the problem will become a piece of cake!