# sequence, convergence

• Jan 14th 2009, 03:15 PM
poincare4223
sequence, convergence
Suppose that $\displaystyle (x_n)$ is a sequence of real numbers. Define a sequence $\displaystyle (y_n)$ by $\displaystyle y_n=\frac{x_n+x_{n+1}}{2}$ $\displaystyle \forall n \in \mathbb{N}$
(a) Prove that $\displaystyle (y_n)$ converges to a real number $\displaystyle x$ if $\displaystyle (x_n)$ converges to $\displaystyle x$.
(b) If $\displaystyle (y_n)$ converges, does $\displaystyle (x_n)$ also converge?
• Jan 14th 2009, 04:02 PM
Jester
Quote:

Originally Posted by poincare4223
Suppose that $\displaystyle (x_n)$ is a sequence of real numbers. Define a sequence $\displaystyle (y_n)$ by $\displaystyle y_n=\frac{x_n+x_{n+1}}{2}$ $\displaystyle \forall n \in \mathbb{N}$
(a) Prove that $\displaystyle (y_n)$ converges to a real number $\displaystyle x$ if $\displaystyle (x_n)$ converges to $\displaystyle x$.
(b) If $\displaystyle (y_n)$ converges, does $\displaystyle (x_n)$ 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

$\displaystyle x_n = \frac{1}{n} + (-1)^n,\;\;\;y_n = \frac{1}{2} \left( \frac{1}{n} + \frac{1}{n+1} \right)$
• Jan 14th 2009, 04:19 PM
ThePerfectHacker
Another counter example for (b) is $\displaystyle x_n = (-1)^n$.
• Jan 14th 2009, 05:01 PM
Jester
Oh Ya - much easier. (Nod)
• Jan 14th 2009, 05:06 PM
poincare4223
I figured it out :-)
• Jan 14th 2009, 07:30 PM
Mathstud28
Quote:

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

For the first one we can easily prove that if $\displaystyle x_n\to x$ then $\displaystyle x_{n+1}\to x$

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

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