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Limit of nrn
By David Loeffler on Monday, April 12, 1999 - 10:14 pm:

While playing around with infinite series I recently came across a problem that has me utterly stumped. I am trying to evaluate the limit, as n tends to infinity, of nrn if r is a constant with modulus less than 1. Intuition and numerical examples suggest that the limit should be 0 but I cannot manage to prove this.

By Dave Sheridan (Dms22) on Tuesday, April 13, 1999 - 12:18 am:

Hi David,

We only need to look at the modulus of the sequence, so I'll assume that r > 0 and leave you to work out how to fill in the rest.

Take logarithms to get n×log(r)+log(n). Now since r < 1 we have that log(r) = -c for some constant c > 0. To recap then, log(nrn)=log(n)-cn.

Please let me know if the rest of this doesn't make sense; I don't know how much background knowledge you have.

Differentiate log(n)-cn to get n-1-c. For large enough n, this is smaller than -c/2. Since y=-x*c/2 decreases to minus infinity as x tends to infinity, so must our function.

So log(nrn)®-¥ which means that nrn®0 as n®¥. Again, if you require details please let me know.

The same method easily extends to showing that p(n) rn®0 for any polynomial p(n) - try adapting the proof to handle this.

Hope this has been helpful.

-Dave

By David Loeffler (Loefflerd) on Tuesday, April 20, 1999 - 11:02 pm:

Thanks - I think I understand most of this, although I will have to go through it a few times. If I get really stuck I will get back to you.

Thanks,

David Loeffler

By Dave Sheridan (Dms22) on Tuesday, April 20, 1999 - 11:06 pm:

Glad you find it helpful. Please don't hesitate to ask about anything you don't understand after having a scribble on some paper - that's by far the best way to understand what's going on.

-Dave

By David Loeffler (P865) on Wednesday, April 21, 1999 - 09:16 pm:

Having read through your argument, the main thing I do not understand is the statement that if f'(x) tends to infinity as x tends to infinity then f(x) does as well. This seems intuitively obvious but is it provable? I attempted a proof by L'Hopital's rule but I could not get anywhere. (Or am I totally barking up the wrong tree?)

-David Loeffler

By Dave Sheridan (Dms22) on Wednesday, April 21, 1999 - 10:51 pm:

I'm not sure quite how l'Hopital can be used here - it's only useful when you explicitly have 0/0, rather than some other way to get infinity. So perhaps you should be finding a different cat.

The result I'm trying to state is much stronger than your question. Here it is:

If f'(x)>c for all x>x0 (a constant) then f(x)®¥ as x®¥.

It doesn't matter what x0 is - we could always shift the graph along a bit and this would not affect the limit. So I'll assume it's 0.

First, imagine a straight line: y=cx. This clearly tends to infinity. Good start.

Now for very small x, f(x)=xf'(d) for some 0<d<x. This is a version of Taylor's theorem, which you might not have come across. If not, you may wish to just say that it's roughly equal to the average gradient between 0 and x. If neither of these convinces you, let me know and I'll try to explain it better.

But we know that f'(d)>c so that f(x)>cx. This only holds for small x though.

Now, take any point z, and assume that f(z)>cz. For small x, f(z+x)=f(z)+f(d) where again 0<d<x. The x and d here may not be the same as the paragraph before, but I'm running out of good letters to use. So again we have f(z+x)>cz+xc=c(z+x).

This is a bit like induction really. It's true around zero and if it's true for some z then it's true for a little larger z. Actually, there's a bit more to it than that - you have to check that "small" means what we want it to. It does of course. That's slightly more complicated.

So if you believe what I've said so far, you should hopefully be convinced that for each x, f(x)>cx and so f(x)®¥.

If you remain unconvinced, it's probably my fault for not explaining it well enough, so don't hesitate to tell me!

-Dave

By David Loeffler (P865) on Monday, April 26, 1999 - 10:56 pm:

Thanks very much, but I'm afraid I don't understand much of your last reply. However, while messing around a little with the problem in the form you stated it, I came across a proof using the mean-value theorem ( (f(b)-f(a))/(b-a)=f'(d), for d between a and b). Is this what you are referring to when you mention a version of Taylor's theorem? If so, I am confused, since you state that it is only valid for small intervals, but I thought the mean-value theorem was always valid for differentiable functions. Anyway, this is what I have come up with:

(f(b)-f(a))/(b-a)=f'(d) (standard MVT)
But f'(d)>c, so (f(b)-f(a))/(b-a)>c. Now let a=0, and assume f(0) = 0. This gives
f(b)>cb. This clearly tends to infinity as b (x?) tends to infinity.

Is this valid, and if not,why not?

Thanks again.

-David Loeffler

By Dave Sheridan (Dms22) on Monday, April 26, 1999 - 11:18 pm:

Sigh! I've been doing things by far too complicated a method. You are, of course, correct. I'd completely forgotten about the MVT. It's not quite the same as Taylor's theorem, but it does have the same consequence here. So ignore my last post and I'll write out your proof in a slightly nicer form.

(f(x)-f(0))/x=f'(d) for some 0<d<x (a way of writing the MVT with a=0). In which case, f(x)>f(0)+cx for every x>0, using the condition f'(d)>c, as you said. So f(x)->infinity as x->infinity. QED.

Thanks for giving me an easy way out of that one! If you're stuck on anything else, or wish to take this discussion in any direction, please let me know. It's been enjoyable so far.

-Dave

By David Loeffler (P865) on Thursday, April 29, 1999 - 10:31 pm:

Thanks very much for "cleaning" my proof there - the original version was probably rather obscure.
Are there any easily comprehensible books I could get hold of that would cover this sort of field? Most of my knowledge of these matters is actually gathered from my dad's old revision notes from a maths course he did as part of his chemistry degree.
If I come across any more problems of this sort I will get back to you. Thanks again.

-David Loeffler

By Dave Sheridan (Dms22) on Tuesday, May 4, 1999 - 10:10 am:

Hi David,

Sorry for not replying sooner. I've been off finding out which books are recommended and useful for this sort of thing (by the Maths dept at Cambridge).

The whole area of limits, sequences and convergence is part of analysis, and the following books are recommended for first year undergraduates:

"A first course in Mathematical Analysis" J C Burkill, Cambridge Uni Press 1978

"Introduction to Mathematical Analysis" J B Reade, Oxford Uni Press (Out of print)

"Calculus" M Spivak, Addison-Wesley/Benjamin-Cummings 1967

Finally, this book was recommended as well, being in "a modern layout" which some of the current first years thought was really good - might be worth checking out.

"Limits - A New Approach to Real Analysis" by A Beardon, Springer 97

Often the differences between these books is the style they're presented in; some seem really interesting and others slightly more dull. Unfortunately, what one person thinks of as good, another finds rather annoying. So it would be best if you could have a look at any of these before buying them - and that can be really difficult. If you're near a city with a university, you might be able to find some of them at the bookshops there. Just a thought.

Hope that's useful for you. If you have any other queries on books (or anything) then let me know. Especially if you can't find them before buying - I can try to help you decide which is the best book for you (but that requires a lot of info I'm afraid...)

-Dave

By David Loeffler (P865) on Wednesday, May 12, 1999 - 10:36 pm:

Thanks - I'm in Bristol, so I'll have a look round the university bookshops. Thanks again for all your help throughout this discussion.

-David Loeffler