Investigating a limit

  • #1
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Homework Statement


Identify the following limits. Indicate if they do not exist. Assume ##r\ne 0##.
##\displaystyle {\lim_{n\to\infty}}(-1)^n( r^n-r^{-n})##
##\displaystyle {\liminf_{n\to\infty}}(-1)^n( r^n-r^{-n})##
##\displaystyle {\limsup_{n\to\infty}}(-1)^n( r^n-r^{-n})##

Homework Equations




The Attempt at a Solution


I'm not sure how I should begin to determine what this limit is, and what the limsup and liminf are.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Identify the following limits. Indicate if they do not exist. Assume ##r\ne 0##.
##\displaystyle {\lim_{n\to\infty}}(-1)^n( r^n-r^{-n})##
##\displaystyle {\liminf_{n\to\infty}}(-1)^n( r^n-r^{-n})##
##\displaystyle {\limsup_{n\to\infty}}(-1)^n( r^n-r^{-n})##

Homework Equations




The Attempt at a Solution


I'm not sure how I should begin to determine what this limit is, and what the limsup and liminf are.
So, what have you tried?
 
  • #3
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So, what have you tried?
What I feel like I should do is first look at limsup and liminf. If they are the same, then that is lim, and if they are different, then lim does not exist.
I have tried to use the definition of limsup directly, where we look at the limit of the supremum of the tails of the sequence. But this doesn't seem to help much... It's just that I am familiar with proving things with limsup and liminf when I am proving general theorems with no concrete sequence. But I have a hard time figuring how to approach specific examples, except when it's obvious, like limsup(-1)^n = 1
 
  • #4
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I would first deal with ##r=1## and then show that we can assume ##r>1## without loss of generality. This gives you some conditions to work with.
 
  • #5
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I would first deal with ##r=1## and then show that we can assume ##r>1## without loss of generality. This gives you some conditions to work with.
That helps. So what it seems like is that if ##0<|r|\le 1##, then we have that ##\displaystyle {\lim_{n\to\infty}}(-1)^n( r^n-r^{-n}) = {\limsup_{n\to\infty}}(-1)^n( r^n-r^{-n}) = {\liminf_{n\to\infty}}(-1)^n( r^n-r^{-n}) = 0##

However, I am not quite sure why we can suppose that ##r>1## WLOG. If we have ##r<-1##, then ##r=-k## for some ##k>1##. Then ##(-1)^n((-k)^n-(-k)^{-n}) = (-1)^n((-1)^n(k)-(-1)^n(k)^{-n}) = (k^n - k^{-n})##, and so we lose the sign.
 
  • #6
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If ##0<|r|<1## then ##0<|r^n|<1## and ##1<|r^{-n}|##. The sequences are symmetric to ##r=1##. That's why you can choose one side, e.g. ##r>1##. Write a few ##a_n## for ##r=2## to see where you have been mistaken. And the case of ##r=1## is somehow easy to deal with first.

Your second argument is basically correct, even if written wrong. We have ##a_n(r)=-a_n(-r)=a_{n+1}(-r)## so the limits are the same, just shifted by a position which doesn't play a role for ##n \to \infty##.
 
  • #7
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If ##0<|r|<1## then ##0<|r^n|<1## and ##1<|r^{-n}|##. The sequences are symmetric to ##r=1##. That's why you can choose one side, e.g. ##r>1##. Write a few ##a_n## for ##r=2## to see where you have been mistaken. And the case of ##r=1## is somehow easy to deal with first.

Your second argument is basically correct, even if written wrong. We have ##a_n(r)=-a_n(-r)=a_{n+1}(-r)## so the limits are the same, just shifted by a position which doesn't play a role for ##n \to \infty##.
Okay, I think I got it. Now, assuming that ##r>1##, it's clear that the ##(-1)^nr^n## dominates, so ##{\limsup_{n\to\infty}}(-1)^n( r^n-r^{-n}) = \infty##, ##{\liminf_{n\to\infty}}(-1)^n( r^n-r^{-n}) = -\infty##, and thus ##\displaystyle {\lim_{n\to\infty}}(-1)^n( r^n-r^{-n}) ## does not exist.

Is this right? And if so, is it at all rigorous? What should be written to make it more rigorous?
 
  • #8
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Okay, I think I got it. Now, assuming that ##r>1##, it's clear that the ##(-1)^nr^n## dominates, so ##{\limsup_{n\to\infty}}(-1)^n( r^n-r^{-n}) = \infty##, ##{\liminf_{n\to\infty}}(-1)^n( r^n-r^{-n}) = -\infty##, and thus ##\displaystyle {\lim_{n\to\infty}}(-1)^n( r^n-r^{-n}) ## does not exist.

Is this right? And if so, is it at all rigorous? What should be written to make it more rigorous?
Yes. But the preliminaries: case ##r=1##, why ##r>0## and then ##r>1## can be considered w.l.o.g. have to be done before.
 
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  • #9
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Yes. But the preliminaries: case ##r=1##, why ##r>0## and then ##r>1## can be considered w.l.o.g. have to be done before.
Actually, I think I am still a bit confused about why we can only consider ##r>1##. What I see is that it's clear that if ##0<|r|\le 1##, we have ##\displaystyle \lim_{n\to\infty}(-1)^n(r^n-r^{-n})=0##, by geometric sequences and the fact that ##(1)^n-(1)^{-n} = 0##. So how can I show that the case ##r<-1## is identical to the case ##r>1##? I see that you wrote ##a_n(r)=-a_n(-r)=a_{n+1}(-r)## but I'm not sure why it's true.
 
  • #10
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Actually, I think I am still a bit confused about why we can only consider ##r>1##. What I see is that it's clear that if ##0<|r|\le 1##, we have ##\displaystyle \lim_{n\to\infty}(-1)^n(r^n-r^{-n})=0##, by geometric sequences and the fact that ##(1)^n-(1)^{-n} = 0##.
This is wrong. For ##r=1## all three limits are zero, yes. But for ##0<r<1## you get ##\lim_{n\to \infty}r^n= 0##, however, ##\lim_{n\to \infty}r^{-n}= \lim_{n\to \infty}\dfrac{1}{r^n}= \infty##
So how can I show that the case ##r<-1## is identical to the case ##r>1##? I see that you wrote ##a_n(r)=-a_n(-r)=a_{n+1}(-r)## but I'm not sure why it's true.
You are right, I made a mistake. I distributed ##(-1)## out while it had to be ##(-1)^n##. So the correction goes for ##r>0\, :## ##a_n(r)=(-1)^n( r^n-r^{-n})=(-1)^n\cdot (-1)^n\cdot( (-r)^{n}-(-r)^{-n})=(-1)^na_n(-r)##. Therefore ##a_n(-r)## alternates differently as ##a_n(r)## does, but this doesn't change the limits, which we consider: ##\lim_{n \to \infty}a_n(r) = \pm \infty## and so is ##\lim_{n \to \infty}a_n(-r) = \lim_{n \to \infty}(-1)^n a_n(r) = \pm \infty##. And even if the sign should cancel out, we still have infinity, either positive or negative for one of them. So maybe the result will get a bit more complicated like superior infinty, and inferior finite, resp. the other way around. (I'm a bit dizzy with all the signs at the moment.)
 
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  • #11
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This is wrong. For ##r=1## all three limits are zero, yes. But for ##0<r<1## you get ##\lim_{n\to \infty}r^n= 0##, however, ##\lim_{n\to \infty}r^{-n}= \lim_{n\to \infty}\dfrac{1}{r^n}= \infty##
You are right, I made a mistake. I distributed ##(-1)## out while it had to be ##(-1)^n##. So the correction goes for ##r>0\, :## ##a_n(r)=(-1)^n( r^n-r^{-n})=(-1)^n\cdot (-1)^n\cdot( (-r)^{n}-(-r)^{-n})=(-1)^na_n(-r)##. Therefore ##a_n(-r)## alternates differently as ##a_n(r)## does, but this doesn't change the limits, which we consider: ##\lim_{n \to \infty}a_n(r) = \pm \infty## and so is ##\lim_{n \to \infty}a_n(-r) = \lim_{n \to \infty}(-1)^n a_n(r) = \pm \infty##. And even if the sign should cancel out, we still have infinity, either positive or negative for one of them. So maybe the result will get a bit more complicated like superior infinty, and inferior finite, resp. the other way around. (I'm a bit dizzy with all the signs at the moment.)
How does this argument look?

Let ##a_n=(-1)^n( r^n-r^{-n})##. We split this into many cases depending on ##r##.

1) First let ##r>1##. Then the limit ##\displaystyle {\lim_{n\to\infty}}(-1)^n( r^n-r^{-n})## does not exist, since the ##r^{-n}## terms goes to zero while the ##r^n## grows exponentially. However, we have the ##(-1)^n## term attached, so in all the limit does not exist but oscillates to infinity in both directions. Hence we also see that ##\limsup_{n\to\infty}a_n = +\infty## and ##\liminf_{n\to\infty}a_n = -\infty##.

2) Second, let ##r<0##. To simplify our sequence, let ##p=-r## so that ##p>0##. Then $$a_n = (-1)^n( r^n-r^{-n}) = p^n-p^{-n}.$$ We see that the ##p^{-n}## term goes to 0 while ##p^n## grows without bound. Hence ##\lim_{n\to\infty}a_n = +\infty##, and also ##\limsup_{n\to\infty}a_n = \liminf_{n\to\infty}a_n = +\infty##.

3) If ##r=1##, then ##(-1)^n(r^n-r^{-n}) = (-1)^n(1-1) = 0##. So ##\lim_{n\to\infty}a_n = \limsup_{n\to\infty}a_n = \liminf_{n\to\infty}a_n = 0##.

4) The last case is ##0<r<1##. Note that ##\frac{1}{r} > 1##. Hence, we have a case roughly identical to the ##r>1## case: ##r^n## goes to zero while ##r^{-n}## grows without bound but also oscillates, so ##\lim_{n\to\infty}a_n## does not exist while ##\limsup_{n\to\infty}a_n = +\infty## and ##\liminf_{n\to\infty} a_n = -\infty##.
 
  • #12
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Looks good, except of part 2. If ##-1<r<0## then the limits will be minus infinity. Also ##r=-1## has a finite value.
 
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  • #13
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Looks good, except of part 2. If ##-1<r<0## then the limits will be minus infinity. Also ##r=-1## has a finite value.
So if I insert these two steps will I be good?
If ##-1<r<0##, then if ##p=-r##, we once again have that ##a_n = p^n-p^{-n}##. But note that ##\frac{1}{p}> 1##. So ##p^n## grows without bound and ##-p^{-n}## goes to negative infinity. So ##\lim_{n\to\infty}a_n = \limsup_{n\to\infty}a_n = \liminf_{n\to\infty}a_n = -\infty##.

If ##r=1##, then ##(-1)^n(r^n-r^{-n}) = (-1)^n(1-1) = 0##. So ##\lim_{n\to\infty}a_n = \limsup_{n\to\infty}a_n = \liminf_{n\to\infty}a_n = 0##.
 
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  • #14
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Yes, that's o.k., maybe some cases can be handled in only one step, as ##|r|=1##, ##|r|<1## and ##|r|>1## but that depends on your preferred style.
 
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