How do I prove that the inductive sequence $y_{n+1}= \dfrac {2y_n + 3}{4}$ is bounded . $y(1)=1$ How do I prove  that the inductive sequence $y_{n+1}= \dfrac {2y_n + 3}{4}$ is bounded? $y(1)=1$
Attempt:  Let us assume that the given sequence is unbounded. : 
Then, $y_{n+1} \rightarrow \infty$ either for some finite $n$ or when $n \rightarrow \infty$

CASE $1 :$ When $|y_n| \rightarrow \infty$ at a finite $n$

Since : $y_{n}= \dfrac {2y_{n-1} + 3}{4}$, then $y_n \rightarrow \pm \infty \implies y_{n-1} \rightarrow  \pm \infty \implies y_{n-2} \rightarrow  \pm \infty ~~\cdots$
This ultimately means $y_2 \rightarrow  \pm \infty$ which is not true.
Hence, there does not exist a finite $n$ for which $y_n \rightarrow  \pm \infty$.

CASE $2:$ When $|y_n| \rightarrow \infty$ at $n \rightarrow \infty$

I think we can proceed the same way as we did above, i.e inductively, we proceed like above and deduce that if the above assumption is true, then $y_2 \rightarrow \infty$, which is not true.
Is my attempt correct?
Does there exist a proof without induction as well?
Thank you for your help.
 A: It is clear that $y_n\gt 0$ for all $n$. Let us prove by induction that $y_n\lt 4$. This is certainly true if $n=1$. And if $y_n\lt 4$, then $\frac{2y_n+3}{4}\lt 4$. 
There are smaller upper bounds than $4$, but to prove boundedness we need not find the least upper bound.
A: An unbounded sequence need not diverge to $\infty$, even if its values are positive.

If the sequence converges, then its limit $l$ satisfies
$$
l=\frac{2l+3}{4}
$$
so $l=3/2$. We could try proving that $y_n\le 3$ for all $n$. The base case is clear, so assume that $y_n\le 3$ and try to prove that $y_{n+1}\le 3$ without peeking below.

 We have $2y_n+3\le 2\cdot3+3=9$, which implies $y_{n+1}\le\dfrac{9}{4}<3.$

The bound below is also easy: $y_n>0$, for all $n$.
A: $$|y_{n+1}-y_n|=\frac{1}{2}|y_n-y_{n-1}|=\frac{1}{2^n}|y_0-y_1|$$
then 
$$|y_n-y_{n+r}|\leq |y_n-y_{n+1}|+|y_{n+1}-y_{n+2}|+...+|y_{n+r-1}-y_{n+r}|=\frac{1}{2^n}|y_0-y_1|\sum_{k=0}^r\frac{1}{2^k}=\frac{1}{2^n}\underset{n\to\infty }{\longrightarrow } 0$$
Then $(y_n)$ is a Cauchy sequence, and so it converge. You can conclude that $(y_n)$ is bounded.
