
Feb 13th, 2012, 09:39 PM
#1
Infinite series
Construct an infinite series of positive terms where sum_{i} a_{i} is finite yet sum_{i} a_{i}^{p} is infinite for every exponent 0 < p < 1. I've constructed one, though it's rather ugly. I was hoping there's a simple example I'm missing.
A simple nearexample is a_{i} = 1/i. The sum diverges for 0 < p <= 1 (from the pseries test, you could say), which is just a little too much divergence. I tried a number of things to modify this series to get convergence at p=1 without messing up the rest, though I had no luck.
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Bertrand Russell
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Feb 23rd, 2012, 10:23 AM
#2
Re: Infinite series
Just to clarify, your use of "i" in your post has nothing to do with the imaginary number i, correct?
On that note, are you trying to restrict yourself to real numbers?

Feb 23rd, 2012, 04:11 PM
#3
Re: Infinite series
Correct, "i" has nothing to do with the imaginary unit. It is an arbitrary index variable denoting a natural number (in my usage, 1, 2, 3, ...). It could be replaced with n throughout. I am restricting the question to real numbers, which is (implicitly) contained in the phrase "series of positive terms."
Here's my solution:
Let S_{p,q} = sum_{n} (1/n^{q})^{p}. From the pseries test, this sum diverges to infinity if and only if 0 <= pq <= 1. Now let q(n) = 1 + 1/n; note that 1q(n) > 1, so each c(n) = S_{1,q(n)} is finite (and obviously nonzero).
Now let
S _{p} = sum _{n} sum _{m} [2 ^{n}/c(n) * 1/m ^{q(n)}] ^{p}
where n and m both range from 1 to infinity. We have cleverly arranged S_{1} to be finite:
S_{1} = sum_{n} sum_{m} 2^{n}/c(n) * 1/m^{q(n)}
= sum_{n} 2^{n}/c(n) sum_{m} 1/m^{q(n)}
= sum_{n} 2^{n}/c(n) S_{1,q(n)}
= sum_{n} 2^{n}/c(n) c(n)
= sum_{n} 2^{n}
= 1
However, for 0 < p < 1, 1/p > 1, so there is some N large enough such that 1/p > q(N) = 1 + 1/N. Thus S_{p,q(N)} is infinite; it follows that S_{p} is infinite:
S_{p} = sum_{n} sum_{m} [2^{n}/c(n) * 1/m^{q(n)}]^{p}
>= sum_{m} [2^{N}/c(N) * 1/m^{q(N)}]^{p}
= [2^{N}/c(N)]^{p} sum_{m} [1/m^{q(N)}]^{p}
= [2^{N}/c(N)]^{p} S_{p,q(N)}
= infinity
Strictly speaking I haven't given the sum in the form I asked for since I used a double sum. This is not a real difficulty, though. It's a basic theorem that rearranging the terms of an infinite series of positive terms does not change the value of the series. We can count off the terms a(n,m) = [2^{n}/c(n) * 1/m^{q(n)}] as, say,
b(1) = a(1,1)
b(2) = a(1,2)
b(3) = a(2,1)
b(4) = a(1,3)
b(5) = a(2,2)
b(6) = a(3,1)
b(7) = a(1,4)
...
The sequence b(i) contains precisely the terms of a(n,m). In light of the above, the sum
has the required properties.
Last edited by jemidiah; Feb 23rd, 2012 at 04:15 PM.
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Feb 23rd, 2012, 04:43 PM
#4
Re: Infinite series
The use (and I agree with your justification... countably infinite sets and all that) of a double summation was a clever step. Very nice.
The reason I asked about the restriction over the reals is that I was considering trying to construct a function over complex numbers whose output always produced positive real numbers, thus falling within the requirement of having only positive terms in the summation.
I'm sure such an approach could construct a solution to your problem (Edit: duh... any solution over the reals IS a solution over the complex numbers), but since you already have your solution, I'll save myself the headache I'm sure I would get trying to work this out.

Feb 23rd, 2012, 07:10 PM
#5
Re: Infinite series
Originally Posted by Lenggries
I was considering trying to construct a function over complex numbers whose output always produced positive real numbers
Of course, there are many, many such functions, eg. f(x + iy) = x^2 + y^2 + 1. If you were at all curious, the problem I was actually working on is more general and at least involves a complexvalued function. I'm not sure if you're familiar with Lebesgue integration, but the statement is essentially, "Suppose f is a complex measurable function on a measure space (X, M, u) where u is a positive measure. Let phi(p) = integral over X of f^p du for 0 < p < infinity. Let E be the set of p where phi(p) is finite. Show that E can be an arbitrary connected subset of (0, infinity)." I specialized it to sequence spaces (so u = counting measure, X = naturals, integral becomes sum) so that people with experience only in infinite sums might have something to say.
My real solution is actually a bit different from what I wrote, though I don't particularly like either example. I constructed a countable disjoint union measure space and a sort of change of variables operation which allowed me to combine E's by intersecting countably many of them. Starting with the sum 1/n series then gives the desired conclusion quickly. I don't like either solution that wellboth are a bit ugly and tedious to write up, though the disjoint union one is perhaps less so since at least one gets a general construction out of the deal.
but since you already have your solution, I'll save myself the headache
If you ever want to spend a few more minutes at it, please feel free. Somehow I have the feeling that I'm ignoring a relatively simple example. Just writing the terms of my example explicitly is extremely messy.
The time you enjoy wasting is not wasted time.
Bertrand Russell
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Feb 24th, 2012, 05:05 PM
#6
Re: Infinite series
Calculus isn't exactly my focus of interest, but the first thing I thought of after reading your last post was the Normal curve. When p=1, the area of the normal curve (aka the probability density function) from negative to positive infinity equals 1 (or equals 1/2 if we only consider positive values of x... not sure where the restriction of "positive measure" in your description applies). I suspect, but am not sure how to prove, that for 0 < p < 1, the integral would be infinite.
Anyway, like I said, calc is not my specialty, but maybe examining probability functions could get you somewhere.

Feb 24th, 2012, 06:31 PM
#7
Re: Infinite series
Unfortunately the normal distribution doesn't work. It dies off very quickly, so it's "hard" to get it to diverge. Thanks for the thought though.
The time you enjoy wasting is not wasted time.
Bertrand Russell
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Mar 1st, 2012, 01:40 PM
#8
Re: Infinite series
OK, I had another thought. In your OP you mentioned the Harmonic series, which you claim has the property of diverging IFF 0 < p <= 1. As I see it (and bear in mind I'm kind of groping in the dark here), for any infinite series with exclusively positive terms, there is some vertex v such that the series diverges IFF 0 < p <= v OR 0 < p < v
This vertex v is the boundary between exponents p, and the boundary can be either inclusive (0 < p <= v) or exclusive (0 < p < v) of the range of values [I]p[/p] which cause the function to diverge.
The function you seek to construct requires an exclusive v such that v = 1.
Now then, my question is this, if you can find a function (any function), which has this exclusive property, such that v is some positive real number, could you perform a transform on the function such that the new v' = 1?
Could it be this simple?
0 < p < v
0 < p/v < 1
v' = p/v
Thus, if you can construct any exclusive series and find the vertex v then all you need to do is raise each term of the series to the 1/vth power.
Is there any merit in this approach, or am I way out of my element?

Mar 1st, 2012, 02:44 PM
#9
Re: Infinite series
Yes, if the series converges for some p, then it converges for all larger p (from the comparison test, since eventually the terms must be less than 1), so the infimum v of the values p for which it converges exists (assuming the series converges for some p somewhere) and is a "boundary" inasmuch as all smaller values diverge and all larger values converge. The series 1+1+1+... shows that v=infinity is entirely possible. I should have said "nonnegative" instead of "positive", since in that case the allzero series 0+0+0+... has v=0. The series with terms e^(n) also works in this vein to give v=0. These two cases are not terribly interesting since they're so easy to dispose ofthe heart of the matter is the 0 < v < infinity case.
Apart from the v=infinity and v=0 cases, yes, you can always transform a given series to an essentially equivalent one with v=1 by taking the vth root of each term. I specialized the problem to v=1 to avoid having an extra variable floating about unnecessarily and to show that I wasn't interested in the v=infinity or 0 cases. Perhaps I should have mentioned that I don't particularly care if 1 is replaced by any other positive real number.
The time you enjoy wasting is not wasted time.
Bertrand Russell
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Mar 1st, 2012, 02:52 PM
#10
Re: Infinite series
Originally Posted by jemidiah
Perhaps I should have mentioned that I don't particularly care if 1 is replaced by any other positive real number.
Yes, that might have been helpful
I work more in theory than in practice, so the problem of specifying v = 1 is that it creates a symptom of target fixation (ie, I get so focused on the 1 that I ignore the larger implications of the problem). Still, I'm glad to see I was able to dust off my memories enough to think through this.
So then your problem now is to come up with any exclusive (my parlance... is there a better word?) series with 0 < v < INFINITY, correct?

Mar 1st, 2012, 04:16 PM
#11
Re: Infinite series
Originally Posted by Lenggries
So then your problem now is to come up with any exclusive (my parlance... is there a better word?) series with 0 < v < INFINITY, correct?
Yes.
As for a better word, I might go with "open" or "closed" to specify p=v diverges or converges, respectively. The motivation is this: in the first case, the set of p's where the series converges is (v, infinity), and in the second case it is [v, infinity). (v, infinity) is an open set but not a closed set while [v, infinity) is a closed set but not an open set, in the topological sense. [That is, [v, infinity) contains all of its limit points while (v, infinity) does not.]
Alternatively you could just say "this series converges at v", "this series converges at the infimum of convergent powers", etc.
The time you enjoy wasting is not wasted time.
Bertrand Russell
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Mar 1st, 2012, 04:34 PM
#12
Re: Infinite series
AHA!!! I had been scratching my head how to notate open and closed sets... glad you mentioned the [ and ( notations.
OK, so this leads me to two questions:
1. I realize that there are infinite many convergent series, so using the phrase "proportion" needs to be taken with a huge grain of salt, but given a random infinite series of the sort we've been discussing, is there any known probability or rough estimate as to whether that series would be open or closed?
2. For a given infinite series, is there a known method for determining v?
I'm guessing the answer to #2 is no, except for maybe in a few special cases (ie the harmonic series). But #1 seems like a fascinating question to me, the kind for which there might be some centuries old unproven conjecture.
BTW, what is the purpose of this search?

Mar 1st, 2012, 05:38 PM
#13
Re: Infinite series
1. Not that I'm aware of. You'd of course need to formalize things moreperhaps by saying the terms are IID random variables taken from [0, 1] uniformly. Of course you'd have to specialize just to the convergent series of this form, since almost all of them would diverge. I don't know enough probability theory to do justice to the question, though. I doubt it's a centuriesold conjecture. I imagine the answer is 1/2:1/2, or possibly 0:1 or 1:0 (though I doubt these two), if it can be formalized at all satisfactorily.
2. I don't know of any nontrivial equivalent definition of v. I suppose you could compute it in binary relatively efficiently by binary search, assuming you can compute convergence or divergence given a particular p. I doubt anything both general and satisfying exists in this vein, though.
The original question is a special case of a part of a problem in Rudin's Real and Complex Analysis, specifically 3.4(c). I'm currently going through the book and doing many of the exercises, including that one.
The time you enjoy wasting is not wasted time.
Bertrand Russell
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Mar 1st, 2012, 05:45 PM
#14
Re: Infinite series
Regarding #1: A binary or similar search would only get you in the neighborhood if the solution space has a smooth topography, which this special set of series should have. Maybe you could email the author for more info on that question... could be an area for an original contribution... or a migraine
Regarding #2: Yeah, even if v could be found, I strongly suspect it would most often be a transcendental number, which means it would be hard to work with from a computer's perspective.

Mar 1st, 2012, 07:01 PM
#15
Re: Infinite series
Well, the coefficients themselves would "usually" be transcendental anyway. That v is often transcendental is pretty much immaterial.
Unfortunately, (Walter) Rudin died a few years ago, though he had a long, presumably happy, life.
The time you enjoy wasting is not wasted time.
Bertrand Russell
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