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$x_n$, $y_n$ be two sequence satisfying $x_n\le y_n\le x_{n+2}$, then which are correct below?

  1. $y_n$ is increasing

  2. $x_n$ and $y_n$ converge together.

  3. $x_n$ decreasing

  4. $y_n$ bounded.

Well, I can see $x_1\le y_1\le x_3$, $x_2\le y_2\le x_4\dots$ by putting $n=1,2..$ but how to compare seperately oftheir individula terms?

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    @BrianM.Scott: Okay, thanks.2012-06-05

2 Answers 2

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From $x_n\le y_n\le x_{n+2}$ for all $n\in\Bbb N$ you can infer that

$x_0\le y_0\le x_2\le y_2\le x_4\le y_4\le\ldots\tag{1}$ and

$x_1\le y_1\le x_3\le y_3\le x_5\le y_5\le\ldots\;,\tag{2}$

but there’s nothing to tie the odd and even subsequences together. In particular, it does not follow that $\langle y_n:n\in\Bbb N\rangle$ is increasing: it might be, for instance, that

$x_n=y_n=\begin{cases} n,&\text{if }n\text{ is even}\\ 1-\frac1n,&\text{if }n\text{ is odd}\;. \end{cases}$

Then the odd and even subsequences are both increasing, but $\langle y_n:n\in\Bbb N\rangle$ is bouncing up and down like crazy. This example also shows that $\langle y_n:n\in\Bbb N\rangle$ need not be bounded (though it can be), and that $\langle x_n:n\in \Bbb N\rangle$ need not be either increasing or decreasing.

I’ll just give you a hint for the remaining one: you can use $(1)$ and $(2)$ to show that if one of $\langle x_n:n\in\Bbb N\rangle$ $\langle y_n:n\in\Bbb N\rangle$ converges, then so does the other, though it’s quite possible that neither converges. Remember, a sequence $\sigma$ converges to some limit $L$ if and only if every subsequence of $\sigma$ converges to $L$.

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The first claim is wrong.

  • For instance, take $x_n = n$ and $y_n = \begin{cases} n+7/4 & \text{if }n \text{ is odd}\\ n + 1/4 & \text{if }n \text{ is even}\end{cases}$

The third claim is obviously wrong.

  • For instance, take $x_n = y_n = n$.

The fourth claim is again wrong.

  • For instance, take $x_n = y_n = n$.

If the second claim is reworded as say they both converge (or) diverge together, then the second claim is right. Else take $x_n = y_n = n$ to see that they both diverge.