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#1
12-21-1999, 02:51 PM
 Boris B Guest Join Date: Jun 1999
A long time ago in math class we were talking about functions which converge at a certain number but never get there. A good example is 1/n. No matter how high n gets, the quotient will never get to zero, but it sure does get close; thus zero is a horizontal asymptote.

Then there was an argument over a different sequence, and I've forgotten how to represent it algebraically. Basically it was 1 plus 1/2 plus 1/3 plus 1/4 etc. I said this didn't converge, it just keeps going higher forever, but I couldn't prove it; others said it does have an asymptote but couldn't tell me what it was. And the mean old teacher didn't clear it up for us! (Actually, I think it was because the bell rang.) So, does it converge when you add a fixed number to the divisor?

(We had already done the 1/2 plus 1/4 plus 1/8 thing and found it converges at 1.)

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- Boris B, Hellacious Ornithologist
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#2
12-21-1999, 03:08 PM
 pluto Guest Join Date: Jul 1999
Way back in my math student days I recall we learned a method to prove whether any sequence converged or not, and how to find the limit of the sequence, etc. However, since the main limit I deal with nowadays is the limit on how much I can remember of what I once knew, I can't say much more than that an answer exists.

If no one else comes up with the answer before tomorrow I'll look it up in the old math book tonight if I can find it.

How's that for a helpful-sounding but ultimately useless answer?

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Let's see. I lathered and I rinsed. But did I repeat?
#3
12-21-1999, 03:26 PM
 Boris B Guest Join Date: Jun 1999
Hey, that's better than my memory. I could barely remember if there was a standard process for determining convergence or not, much less what it was.
#4
12-21-1999, 04:03 PM
 DSYoungEsq Guest Join Date: Jul 1999
Whoa, now I'm having nightmares about college math! THANKS, Boris.

Still, if the sequence of 1/n + 1/(n+2) + ... has a limit, why wouldn't the sequence of 1 + 1/n + 1/(n+1) + ... ? And isn't this basic calculus??
#5
12-21-1999, 04:12 PM
 pluto Guest Join Date: Jul 1999
Ta-da!

The series diverges.

I walked over to the technical library (which is swarming with math books) and looked it up. Hey! It sure beats working.

The first book I found used that series as an example of a tricky one because it passes the "limit test", that is, each term is smaller than the previous one so the limit (not the sum) of the series is zero. This is a necessary condition for convergence but not a sufficient condition!

The series fails the "ratio test" -- the limit of the ratio of successive terms must go to zero. In this case the first few ratios are 1/2, 2/3, 3/4 ..., which converges to 1.

I'm not certain I got the details right. I forgot most of it while walking back to my building even though I took notes (see previous post). But that puppy diverges.

A note in another book said that the sum of the series 1/n^k, with n from 1 to infinity, is known as the Riemann zeta function and it is important in number theory. It converges if k > 1. Your series uses k = 1, which puts it in the divergent realm.

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Let's see. I lathered and I rinsed. But did I repeat?
#6
12-21-1999, 05:11 PM
 Cabbage Guest Join Date: Sep 1999
Here's an easier way to see that this diverges. Group the terms like this:

1 + 1/2 + (1/3 + 1/4) + (1/5 + ... + 1/8) + (1/9 + ... + 1/16) + (1/17 + ... + 1/32) + ...

Following this pattern, the sum inside each set of parentheses will be bigger than 1/2, so the sum just keeps getting bigger without bound.
#7
12-21-1999, 05:38 PM
 pluto Guest Join Date: Jul 1999
Yeah, but then you don't get to use terms like Riemann zeta function.

Oh all right. That's a clever proof.
#8
12-21-1999, 10:52 PM
 Boris B Guest Join Date: Jun 1999
Bravo, Pluto! You're far out!

Bravo, Cabbage! I can't think of good pun based on your name.

I haven't heard the name Riemann in a loooong time.
#9
12-22-1999, 04:07 AM
 bantmof Guest Join Date: Jun 1999
BorisB says:
Quote:
 I said this didn't converge, it just keeps going higher forever, but I couldn't prove it...
You were correct, and here's one way to show it, plus a note on the above mentioned ratio test. For ascii notational convenience let sum(x=n:m,Ux) denote the sum from x=n to m of the series Ux. Let intg(n:m, f(x) dx) denote the integral operator, and let oo denote infinity.

Your series 1/1+1/2+1/3+1/4... is then sum(x=1:+oo, 1/x). From the Cauchy integral test, this series (which meets the required conditions of being decreasing and continuous) converges if and only if intg(1:+oo, dx/x) is finite.

intg(dx/x) = log(x)
log(oo) = oo ==> divergent series.

Also, a note on the ratio test:
Quote:
 The series fails the "ratio test"
I'm not entirely sure you can show this series diverges using this test. I might be wrong here since it's been a while since I've done this, but, the ratio test is formally stated thusly:

If Un is a positive series, and p=lim(n->oo, U(n+1)/Un), then the series converges if p<1, diverges if p>1, and (here's the important bit) may or may not converge for p=1.

In the case of the above series, we get:
p=lim(n->oo, (1/(n+1))/(1/n))
p=lim(n->oo, n/(n+1)) = 1

So the ratio test does not show this series converges (although it allows for it). At least, if I'm doing all this right - it's been more decades since my last math class than I like to count. :-)

--
peas on earth
#10
12-22-1999, 04:10 AM
 bantmof Guest Join Date: Jun 1999
Crap. The ubb codes mangled my ascii notation.

Those red face-things are supposed to be 1 : oo. The integral was supposed to read: intg(1 : oo, dx/x).

--
peas on earth
#11
12-22-1999, 04:15 AM
 bantmof Guest Join Date: Jun 1999
Ok, this is really not my day - somebody just put me out of my misery. Part of this paragraph about the limit test also got eaten by HTML:
Quote:
 If Un is a positive series, and p=lim(n->oo, U(n+1)/Un), then the series converges if p
That should read:

If Un is a positive series, and p=lim(n->oo, U(n+1)/Un), then the series converges if p &lt; 1, diverges if p &gt; 1, and may or may not converge for p=1.

--
peas on earth
#12
12-22-1999, 06:01 AM
 Omniscient Charter Member Join Date: Apr 1999 Location: Chicago, IL, USA Posts: 15,768
Oooo, Dif Eq nightmares. Fuck you man, I need a drink.
#13
12-22-1999, 07:10 AM
 C K Dexter Haven Right Hand of the Master Administrator Join Date: Feb 1999 Location: Chicago north suburb Posts: 14,681
Note that there is a difference between the SEQUENCE { 1/n } converging to an asymptote and the SERIES 1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ... which does not converge.

The series is the sum of elements in the sequence. The first condition for a series to be considered for converging is whether the sequence converges to zero; if the sequence doesn't converge to zero, no way the series will converge.

Thus, the series { 1/n } converges ("gets very close") to zero as an asymptote ("never actually is equal to zero, just gets close.") But the sequence SUM( 1/n ) does not converge.

Another fun example is the sequence 1 - 1 + 1 - 1 + 1 ...

If you group it as (1 -1) + (1 -1) + (1 - 1) + ... it looks like it converges to zero. If you group it as 1 + (-1 +1) + (-1 + 1) + ... you think it converges to 1. In fact, since the underlying series { 1 } doesn't converge to zero, this sequence doesn't converge at all.

Another interesting tidbit for math geeks: Say you have a polynomial fraction. Factor the numerator and the denominator. Consider where a value of x makes one of the factors in the denominator equal to zero. If that factor can be cancelled out (same factor appears in numerator), then the graph of the function has a hole in it (at the point where that factor would be zero.) If that factor cannot be cancelled out, then the graph of the function has an asymptote (at the point where that factor would be zero.) Someone who doesn't know the difference -- are your ready? -- can't tell their asymptote from a hole in the graph.

I thank you.
#14
12-22-1999, 11:36 AM
 DSYoungEsq Guest Join Date: Jul 1999
Yikes!!! I was RIGHT, this IS what made me a Poli Sci major!

<heading after Omniscient>
#15
12-22-1999, 03:19 PM
 torq Guest Join Date: Mar 1999
It's not necessary for the limit of the ratio of successive terms to go to zero for the sum of the series to be convergent. The example 1/2 + 1/4 + 1/8 + 1/16 ...
has a limiting ratio of 1/2, but the sum of the series converges (to 1).
#16
12-22-1999, 03:33 PM
 pluto Guest Join Date: Jul 1999
To save any more of you from correcting me, I realize that I got the ratio test description (and application) screwed up. I told you I forgot it before I got back to my desk.

Sorry.

On the other hand I did correctly identify the series as divergent and that it was tricky, even though I didn't get the reason right. So I'll take partial credit for answering the question. I also found a couple of useful books while I was in the library so it was worth my while in spite of problems with my memory.

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Let's see. I lathered and I rinsed. But did I repeat?
#17
12-22-1999, 05:06 PM
 knappy Guest Join Date: Dec 1999
Speaking of Riemann, here are links to a report of possible progress on the Riemann hypothesis, the biggest unsolved problem in math:
http://www.maa.org/mathland/mathtrek_6_21_99.html
http://www.maa.org/mathland/mathtrek_6_28_99.html
#18
12-23-1999, 08:52 AM
 Guest
torq says: << It's not necessary for the limit of the ratio of successive terms to go to zero for the sum of the series to be convergent. The example 1/2 + 1/4 + 1/8 + 1/16 ...
has a limiting ratio of 1/2, but the sum of the series converges (to 1). >>

Correct.

However...First: It IS necessary for the sequence to converge to zero for the series (sum) to converge. In torq's example, the underlying sequence { 1/n2 } certainly converges to zero.

Second: In terms of the ratio of successive terms, suppose there is a sequence { an }. Take the ratio of successive terms, an+1/an. If that ratio is less than a fixed positve number A, then the series converges if A < 1 and the series diverges if A >= 1. This is commonly called the ratio test and is attributed to Cauchy (1821).

Note that the ratio test does not say that the ratio of successive terms must tend to zero, only that it must be less than a fixed positive number (less than 1.) ((Technical footnote: ratio test really says that this is true AFTER SOME POINT M in the series... the first zillion points might not fit the ratio test, as long as AFTER the zillion points, the ratio test is satisfied, then the series will converge.))

[Note: This message has been edited by CKDextHavn]
#19
12-24-1999, 09:27 PM
 Lumpy Charter Member Join Date: Aug 1999 Location: Minneapolis, Minnesota US Posts: 10,927
Ok, the sum of some series (or is it sequence? Dang, I never took calculus) is convergent, while other series are divergent. Is there a series that is divergent, but only just barely? That's "parabolic", so to speak?
#20
12-24-1999, 10:24 PM
 Guest
Yes, the series { 1/n } diverges "just barely" in the sense that the series { 1/na } converges when a > 1, but diverges at a = 1.

[Note: This message has been edited by CKDextHavn]

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