Convergence and recursive sequences Let $\{y_n\}$ be a sequence such that $y_1 = 1$ and $y_{n+1} = \frac{2y_n +3}{4}$, if $n\geq 1$. I want to determine whether this sequence converges or not. It seems pretty clear to me that this sequence is monotonically increasing and also that $1\leq y_n \leq \frac{3}{2}$ for all $n\in \Bbb{N}$. I couldn't prove the upper bound by induction, so I'm now trying to show that this is a Cauchy sequence and thus converges. Solving for $y_n$, we get $$y_n = \frac{4y_{n+1} -3}{2}$$
Next I tried to show that $\vert y_{n+1} - y_n\vert$ tends to zero as $n\to \infty$, however I ended up with $$\vert y_{n+1} - \frac{3}{2}\vert$$
Which makes sense since the terms are tending to $\frac{3}{2}$ as $n\to \infty$, but I don't know how to develop the idea from here. Is this an induction problem or a Cauchy problem, or something else?
 A: Proving the upper bound is actually quite simple. Notice that if $y_n\le3/2$, then
$$y_{n+1}=\frac{2y_n+3}4\le\frac{2\times\frac32+3}4=\frac32$$
so it holds by induction. Likewise, showing it is increasing simply amounts to showing that
$$\frac{2y+3}4\ge y$$
for any $y\le3/2$. This is clear if we rearrange it:
$$\frac{2y+3}4\ge y$$
$$2y+3\ge4y$$
$$3\ge2y$$
$$\frac32\ge y$$
$$y\le\frac32$$
and so we have
$$y_{n+1}=\frac{2y_n+3}4\ge y_n$$
for all $y_n\le3/2$, which we know is true from above.
A: We can solve that recurrence relation actually. Let $ n $ be a positive integer.
\begin{aligned} \left(\forall k<n\right),\ y_{n-k}&=\frac{1}{2}y_{n-k-1}+\frac{3}{4} \\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\forall k<n\right),\ \frac{1}{2^{k}}y_{n-k}&=\frac{1}{2^{k+1}}y_{n-k-1}+\frac{3}{2^{k+2}}\\ \iff \sum_{k=0}^{n-2}{\left(\frac{1}{2^{k}}y_{n-k}-\frac{1}{2^{k+1}}y_{n-k-1}\right)}&=3\sum_{k=0}^{n-2}{\frac{1}{2^{k+2}}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  y_{n}-\frac{y_{1}}{2^{n-1}}&=3\left(\frac{1}{2}-\frac{1}{2^{n}}\right)\end{aligned}
Thus : $$ \left(\forall n\in\mathbb{N}\right),\ y_{n}=\frac{3}{2}-\frac{1}{2^{n}} $$
