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Suppose that $x_n$ $\rightarrow$ + $\infty$and $y_n$ $\rightarrow$ a

True or false? If a>0, then $x_ny_n$ $\rightarrow$ + $\infty$

This is what I have:

Let $\epsilon$>0 then $n_0$ $\in$ $\mathbb{N}$ for |$y_n$-a|<$\epsilon$. Let $\epsilon$=a/2 then |$y_n$-a| < a/2 so, -a/2< $y_n$-a < a/2 add a to both sides and a/2< $y_n$ < 3a/2. Therefore, $y_n$>a/2 for n> $n_0$ if $y_n$> a/2 and $x_n$ $\rightarrow$ $\infty$. So, $x_ny_n$> $x_n$(a/2) which equals $x_ny_n$> +$\infty$ (a/2)= +$\infty$. Therefore the statement is true.

I was wondering if this would also work if $\epsilon$=a/4 with the proof:

Let $\epsilon$>0 then $n_0$ $\in$ $\mathbb{N}$ for |$y_n$-a|<$\epsilon$. Let $\epsilon$=a/4 then |$y_n$-a| < a/4 so, -a/4< $y_n$-a < a/4 add a to both sides and 3a/4< $y_n$ < 5a/4. Therefore, $y_n$>3a/4 for n> $n_0$ if $y_n$> 3a/4 and $x_n$ $\rightarrow$ $\infty$. So, $x_ny_n$> $x_n$(3a/4) which equals $x_ny_n$> +$\infty$ (3a/4)= +$\infty$. Therefore the statement is true.

or can $\epsilon$ only equal a/2?

  • Just out of curiosity, why did you write some things in $\LaTeX$ and not others? – Clayton Mar 11 '19 at 22:11
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    $\epsilon$ can be any value and for that value you can find an $N$ where $n > N$ will yield $|y_n -a |< \epsilon$ and a value $M$ where $x_n > M$. And really you MUST get out of the habit of treating $\infty$ as though it were a number. – fleablood Mar 11 '19 at 22:11
  • fleablood what do you mean?? – Reign T. Mar 11 '19 at 22:13

2 Answers2

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As noted above, you do have the right idea. Let's try cleaning it up so that you can see how these proofs should be written. We have

$$\lim_{n \to \infty} x_n = \infty \text{ and } \lim_{n \to \infty} y_n = a \gt 0.$$

Choose $K$ arbitrarily large. We need to find some $N$ such that $n \gt N \Rightarrow x_ny_n \gt K.$ If we can always do that for any choice of $K$, then we are done.

Let $\epsilon = a/2 \gt 0$. For some $M_1$, we know $n \gt M_1 \Rightarrow |y_n-a| \lt \epsilon \Rightarrow |y_n| \gt a/2$.

We also know that for some $M_2$, we know $n \gt M_2 \Rightarrow x_n \gt 2K/a$.

Choose $N = \max(M_1, M_2)$. Then $n \gt N \Rightarrow x_ny_n \gt (2K/a)(a/2) = K$ and we are done.

Notice how I never needed to try to manipulate $\infty$ as a number. You choose an arbitrarily large number (I called it $K$), which you are allowed to manipulate, and then wrote a proof that all elements of the sequence of interest past a certain point ($N$) must be even bigger than $K$.

The problem with writing the proof the way you did is that you're essentially assuming the result that you're trying to prove. How do you know that $x_n(3a/4) \rightarrow \infty$ until you've completed this proof?

Robert Shore
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As you have already noted, the same method of proof works for $\epsilon=r\cdot a$ where $r$ is any real number strictly between 0 and 1. Your two choices would correspond to $r=1/2$ and $r=1/4$.

  • so they are both correct? – Reign T. Mar 11 '19 at 22:11
  • $\epsilon$ can be ANY value. You pick one that works. – fleablood Mar 11 '19 at 22:13
  • Your proof needs to be cleaned up a bit (not manipulating infinity as a number, for instance), but the idea is correct. – TM Gallagher Mar 11 '19 at 22:13
  • well 3a/4 goes to infinity as well and I thought when doing limits you could have $\infty$ * $\infty$= $\infty$? – Reign T. Mar 11 '19 at 22:16
  • To rigorously show that the sequence $(x_ny_n)_n$ goes to $\infty$ as $n\to\infty$, you must show that, for all $M>0$, there exists $n_0$ for which $x_ny_n>M$ whenever $n>n_0$. The statement “3a/4 goes to infinity” doesn’t make sense since $a$ is a fixed number and doesn’t go anywhere. – TM Gallagher Mar 11 '19 at 22:26