1. ## sequence, convergence

Suppose that $(x_n)$ is a sequence of real numbers. Define a sequence $(y_n)$ by $y_n=\frac{x_n+x_{n+1}}{2}$ $\forall n \in \mathbb{N}$
(a) Prove that $(y_n)$ converges to a real number $x$ if $(x_n)$ converges to $x$.
(b) If $(y_n)$ converges, does $(x_n)$ also converge?

2. Originally Posted by poincare4223
Suppose that $(x_n)$ is a sequence of real numbers. Define a sequence $(y_n)$ by $y_n=\frac{x_n+x_{n+1}}{2}$ $\forall n \in \mathbb{N}$
(a) Prove that $(y_n)$ converges to a real number $x$ if $(x_n)$ converges to $x$.
(b) If $(y_n)$ converges, does $(x_n)$ also converge?
I think that (a) can be proven fairly easy within the definition of convergent sequences.

(b) on the other hand I think is false. A possible counter example is

$x_n = \frac{1}{n} + (-1)^n,\;\;\;y_n = \frac{1}{2} \left( \frac{1}{n} + \frac{1}{n+1} \right)$

3. Another counter example for (b) is $x_n = (-1)^n$.

4. Oh Ya - much easier.

5. I figured it out :-)

6. Originally Posted by poincare4223
Suppose that $(x_n)$ is a sequence of real numbers. Define a sequence $(y_n)$ by $y_n=\frac{x_n+x_{n+1}}{2}$ $\forall n \in \mathbb{N}$
(a) Prove that $(y_n)$ converges to a real number $x$ if $(x_n)$ converges to $x$.
(b) If $(y_n)$ converges, does $(x_n)$ also converge?
For the first one we can easily prove that if $x_n\to x$ then $x_{n+1}\to x$

Proof: We know that since $x_n\to x$ we may choose $N$ such that $n\leqslant N$ then $d(x_n,x)<\varepsilon$ and since $N\leqslant n\implies N\leqslant n+1$ we must have that $d(x_{n+1},x)<\varepsilon$

This implies that $\lim x_n,\lim x_{n+1}$ and consequently $\lim y_n$ exist. So because of this existence $\lim y_n=\frac{1}{2}\left[\lim x_n+\lim x_{n+1}\right]=\frac{1}{2}[x+x]=x$