How to demonstrate the sequence $y_{n+1}=\frac{n}{n+1} + \frac23 y_n$ converges.

Question

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 $(y_n)$?

Answer 1

$$ y_{n+1} = \frac{n}{n+1} + \frac23 \cdot y_n \tag{2} $$ implies that the sequence is bounded: $$ |y_n| \le \max(3, |y_0|) $$ is easily shown with induction. Therefore both $$ I = \liminf_{n\to \infty} y_n \, , \, S = \limsup_{n\to \infty} y_n $$ are finite, and then $(2)$ implies that $$ I = 1 + \frac 23 I \, , \, S = 1 + \frac 23 S $$ so that $I=S= 3$.

Answer 2

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 $\lim_{n\to \infty } y_n = 3$.

Answer 3

Let $z_n = y_n - 3$. We have $z_{n+1} = \frac23 z_n - \frac{1}{n+1}.$

We use Mathematical Induction to prove that $|z_n| \le \frac{3|z_0| + 12}{n+1}$ for all $n \ge 2$.

When $n = 2$, true.

Assume that it is true for $n$ ($n\ge 2$).

For $n+1$, we have $$|z_{n+1}| \le \frac23|z_n| + \frac{1}{n+1} \le \frac23 \cdot \frac{3|z_0| + 12}{n+1} + \frac{1}{n+1} \le \frac{3|z_0| + 12}{n+2}$$ where we use $$\frac{3|z_0| + 12}{n+2} - \frac23 \cdot \frac{3|z_0| + 12}{n+1} - \frac{1}{n+1} = \frac{(n-1)|z_0| + 3(n-2)}{(n+1)(n+2)} \ge 0.$$

We are done.