Context
While answering this question I noticed that the procedure used there could be extended to show that any sequence of the form
$$y_n = \sum_{k=0}^n a_{n-k}x_k, \ \ n=0,1,2\dots\tag{1}\label{eq1}$$
is null if the series $\sum a_n$ converges absolutely and $(x_n)$ is a null sequence.
I wanted to create some examples, as variations of the one proposed in the above mentioned question, and came up with, e.g.,
$$y_{n+1} = \frac{n}{n+1} + \frac23 \cdot y_n, \ \ n=0,1,2,\dots,\tag{2}\label{eq2}$$
and arbitrary $y_0\neq 0$.
With the change of variable $z_n = y_n -3$ you can then write
$$z_n = \sum_{k=0}^n \left(\frac{2}{3}\right)^{n-k}x_k,$$
with
$$
x_n=
\begin{cases}
z_0 & (x=0)\\
-\frac{1}{n+1} & (n>0),
\end{cases}
$$
that is exactly the form \eqref{eq1}, and it satisfies the hypotheses. So the sequence $(z_n)$ is null and $(y_n)$ converges to $3$, independently of the intial value $y_0$.
Question
I think from \eqref{eq2} the convergence to $3$ is sort of obvious, but I was not able to come up with a less "convoluted" way to show it. Could you provide me with an alternative proof of the convergence of $(y_n)$?
Best Answer
Take a look at the iteration of the form $z_{n+1} = a+\frac{2}{3}z_n$ where $a$ is constant. Then subtracting $3a$ you get $z_{n+1}-3a = \frac{2}{3}(z_n-3a)$. This means that $z_{n}-3a$ is geometric and it always converges to $0$ regardless of the initial value $z_0$. This means that $z_n\to 3a$.
Now it is enough to see that for some $\varepsilon >0$ eventually you have $n/(n+1)>1-\varepsilon$ if $n \geq n_0$. Therefore, you consider the recurrence of the form $z_{n+1} = 1-\varepsilon + \frac{2}{3}z_n$ and $w_{n+1} = 1+\frac{2}{3}w_n$ for $n \geq n_0$ and $z_{n_0}=w_{n_0}=y_{n_0}$ then $ z_n\leq y_n \leq w_n$ for every $n$. The initial paragraph above says that
$$ 3-3\varepsilon \leq \liminf_{n\to \infty} y_n \leq \limsup_{n \to \infty} y_n \leq 3 $$
Now it is enough to take $\varepsilon \to 0$ to get that $\lim_{n\to \infty } y_n = 3$.