# Thread: Sequence Convergence proof maybe?

1. ## Sequence Convergence proof maybe?

a) Suppose that ${X_n}$ and ${Y_n}$ converge to the same point. Prove that ${X_n} - {Y_n} \rightarrow 0$ as $n \rightarrow \infty$.

b) Show the converse is false.

Again as usual... some kind of explanation to help me understand this helps alot...

2. For part a) i would use the $\epsilon-\delta$ definition of limits and it should follow directly. For part b) you just need to show a counter example. Maybe look at $y=x\mbox{ and }y^2-x^2=1.$

3. ok that was how I started it... thanks for the info...

4. Originally Posted by Caity
a) Suppose that ${X_n}$ and ${Y_n}$ converge to the same point. Prove that ${X_n} - {Y_n} \rightarrow 0$ as $n \rightarrow \infty$.
b) Show the converse is false.
Suppose that $L$ is the common limit of the two sequences.
If $\varepsilon > 0 \Rightarrow \quad \left( {\exists N} \right)\left[ {n \geqslant N \Rightarrow \left| {x_n - L} \right| < \frac{\varepsilon }
{2}\,\& \,\left| {y_n - L} \right| < \frac{\varepsilon }
{2}} \right]$
.
Then observe that $\left| {x_n - y_n } \right| \leqslant \left| {x_n - L} \right| + \left| {L - y_n } \right|$.

5. I finished part a). Thanks for the help. Now about part b). First what exactly would the converse be? I know its If X then Y becomes If Y then X. I was thinking maybe by showing something like subtraction is not commutative. Like if X - Y $\rightarrow$ 0 is not the same as Y - X $\rightarrow$0.

6. The converse is: Suppose that $\left( {x_n - y_n } \right) \to 0$ the $\left( {x_n } \right)$ & $\left( {y_n } \right)$ have the same limit.

7. Originally Posted by Plato
The converse is: Suppose that $\left( {x_n - y_n } \right) \to 0$ the $\left( {x_n } \right)$ & $\left( {y_n } \right)$ have the same limit.

Well I tried this Plato... but it comes out true and I'm supposed to be proving it false... If the limit of $x_n$ and $y_n$ go to the same number say 1. Then we have 1 - 1 = 0 which shows it is true.

8. Try this counterexample.
$x_n = \left( { - 1} \right)^n + \frac{1}
{n}\,\& \,y_n = \left( { - 1} \right)^n - \frac{1}
{n}\,$

Note: we are not given that the sequences converge, only that their difference is null.

9. Originally Posted by Plato
Note: we are not given that the sequences converge, only that their difference is null.
Is it possible to find a counterexample with the two converging ?

10. Originally Posted by Moo
Is it possible to find a counterexample with the two converging ?
The answer is no, the limit must be the same for both sequences.
Suppose that $\left( {x_n } \right) \to L,\quad \left( {y_n } \right) \to K\,\& \,\left| {\left( {x_n } \right) - \left( {y_n } \right)} \right| \to 0$.
That means that almost all the terms of $\left( {x_n } \right)$ are ‘close’ to $L$; almost all the terms of $\left( {y_n } \right)$ are ‘close’ to $K$; and almost all the terms of $\left( {x_n } \right) - \left( {y_n } \right)$ are ‘close’ to $0$.

Suppose $L \not= K$ then $\varepsilon = \frac{{\left| {L - K} \right|}}
{4} > 0$
.
But how can almost all of the three sequences be with in an $\epsilon$-distance of its limit?