Questions about infinity

[Kriminal99]

I came across a math problem with a given result that I confusing and that seemed to contradict other things stated as fact within the same source.

 

The book states that the infinite harmonic series 1 + 1/2 +1/3 + 1/4 ... diverges. Even though the limit of the value of term N in the sequence as N approaches infinity is 0. (Is this just because the fractions don't get small enough fast enough, or is this a typo or what?)

 

Then there is a problem that reduces to the following:

 

[math]\displaystyle(\frac{1}{3})\sum_{x=0}^{\infty}[\frac{1}{x+1}-\frac{1}{x+4}][/math]

The answer in the book claims that all that is left from this is 1 + 1/2 + 1/3 because it basically says (1 + 1/2 + 1/3 + 1/4 ...) - (1/4 + 1/5 + ...)

 

But this reasoning entails an overlap of different iterations of the sum that should not be occuring. IE for x = 1, you have 1/2 - 1/4. for x = 3 you have 1/4 - 1/7. Only by using different values of x can you get them to cancel each other out. So if you follow the sum to any definite value of i such as 10,000 there will always be some difference from the result they gave.

 

Of course that difference will go to 0 as x goes to infinity. I would have appreciated an answer that showed this rather than trying to claim that the two series cancel each other out even though for any given value of the dummy variable they do not. Is this actually how most people think infinity works?

 

Also I am confused because the original problem that reduced into the above was:

 

[math]\displaystyle\sum_{x=0}^{\infty}\frac{1}{x^2+5x+4}[/math]

 

So why would this converge any more than a harmonic series, given that in this form it is an infinite series of fractions where the terms just get smaller and smaller?

 

IE this would be the series:

 

(1/4+1/10+1/18....)

[CraigD]

… I would have appreciated an answer that showed this rather than trying to claim that the two series cancel each other out even though for any given value of the dummy variable they do not.

I share your dissatisfaction with this method of presenting the problem. In their effort to encourage intuitive understanding of math problems, textbook authors sometime, I think, fail to also adhere to the formal approach so important in math – though they can always use the excuse that adhering to formalism is “left as an exercise for the reader”. ;)

 

IMHO, this is a more satisfying algebra:

 

[math]\frac13 \sum_{x=0}^{\infty}[\frac{1}{x+1}-\frac{1}{x+4}] =[/math]

 

[math] \frac13 \left( \sum_{x=0}^{\infty}[\frac{1}{x+1}] - \sum_{x=0}^{\infty}[\frac{1}{x+4}] \right) =[/math]

 

[math] \frac13 \left( \sum_{x=0}^2 [\frac1{x+1}] + \sum_{x=3}^{\infty}[\frac1{x+1}] - \sum_{x=0}^{\infty}[\frac1{x+4}] \right) =[/math]

 

[math] \frac13 \left( \sum_{x=1}^3 [\frac1x] + \sum_{x=4}^{\infty}[\frac1x] - \sum_{x=4}^{\infty}[\frac1x] \right) =[/math]

 

[math] \frac13 \sum_{x=1}^3 [\frac1x] = \frac{11}{18}[/math]

Also I am confused because the original problem that reduced into the above was:

 

[math]\displaystyle\sum_{x=0}^{\infty}\frac{1}{x^2+5x+4}[/math]

 

So why would this converge any more than a harmonic series, given that in this form it is an infinite series of fractions where the terms just get smaller and smaller?

Because each term [math]\frac{1}{x^2+5x+4}[/math] in this exactly equals each term [math]\frac13 \left( \frac1{x+1} -\frac1{x+4} \right)[/math] in the previous series. This can be proven by simple algebra, or demonstrated by evaluating both terms for several test values of [math]x[/math].

[Qfwfq]

 

[math]\displaystyle\sum_{x=0}^{\infty}\frac{1}{x^2+5x+4}[/math]

 

So why would this converge any more than a harmonic series

Because the denominator's dominant term is [math]\norm x^2[/math] and not [math]\norm x^1[/math].

 

I share your dissatisfaction with this method of presenting the problem.

I don't. I would criticize the book only if it has fails to first point out that the given series does converge, without which no algebra will strictly be conclusive. Krim's critique of "Only by using different values of x" is irrelevant. If it weren't, one could point out that your algebra really has the same "problem". Consider a different series:

 

[math]\sum_{x=0}^{\infty}2^x[/math]

 

which is clearly divergent. Yet, by algebra and with a cancellation of exactly sums just as in your method, one can "prove" it's value to be -1. In a similar manner one can "prove" the value of:

 

[math]\sum_{x=0}^{\infty}(-2)^x[/math]

 

to be 1/3.

 

Risky to confuse algebra and topology.........

[Pyrotex]

...I don't. I would criticize the book only if it has fails to first point out that the given series does converge, without which no algebra will strictly be conclusive.

What? :confused: I don't understand you. What do you mean?

Krim's critique of "Only by using different values of x" is irrelevant. If it weren't, one could point out that your algebra really has the same "problem".

What? :confused: I don't understand you. What do you mean?

Consider a different series:

 

[math]\sum_{x=0}^{\infty}2^x[/math]

 

which is clearly divergent. Yet, by algebra and with a cancellation of exactly sums just as in your method,

What? :confused: I don't understand you. What do you mean?

[Pyrotex]

...The book states that the infinite harmonic series 1 + 1/2 +1/3 + 1/4 ... diverges. Even though the limit of the value of term N in the sequence as N approaches infinity is 0. (Is this just because the fractions don't get small enough fast enough, or is this a typo or what?)

The book is correct. Just because an infinite series goes to zero does not mean that the sum must converge. Summing up 1/N long enough in this series does indeed surpass any arbitrary value. One might say, that 1/N doesn't approach zero fast enough

....So why would this converge any more than a harmonic series, given that in this form it is an infinite series of fractions where the terms just get smaller and smaller?...

Again, the mere fact that that the terms get smaller and smaller is insufficient to determine whether or not the infinite series sum will approach some particular finite value, or approach infinity. The rate at which the terms decrease is very important.

[Pyrotex]

...[math]= \frac13 \sum_{x=1}^3 [\frac1x] = \frac{11}{18}[/math]....

 

[math]= \frac{1}{3}\sum_{x=1}^3 [\frac1x] = \frac{1}{3} * [\frac{1}{1} + \frac{1}{2} + \frac{1}{3}] = \frac{1}{3} * [\frac{10}{6}] = \frac{5}{9}[/math]

[CraigD]

For a moment I thought I’d had a episode of arithmetic senility, but I think the err is Pyro’s, not mine. ;)

[math]\frac{1}{3} * [\frac{1}{1} + \frac{1}{2} + \frac{1}{3}] = \frac{1}{3} * [\frac{10}{6}] = \frac{5}{9}[/math]

 

[math]\frac13 [ \frac11 + \frac12 + \frac13 ]= \frac13 [ \frac66 + \frac36 + \frac26 ] = \frac13 \times \frac{6 + 3 + 2}6 = \frac13 \times \frac{11}6 = \frac{11}{18}[/math]

[Qfwfq]

I vote for Craig's 11/18!!!!! :D

 

What? :confused: I don't understand you. What do you mean?

I mean that algebra isn't topology. I could also say that algebra including sums that are meaningless without topology to determine that they have a finite value might even be called pseudo-algebra. In the specific case the two sums aren't finite at all, but at least they are exactly equal. This is however a rather hair's breadth argument, without a spot of topology.

 

What? :confused: I don't understand you. What do you mean?

What Craig does, writing the two terms:

 

[math]\sum_{x=0}^2 [\frac1{x+1}] + \sum_{x=3}^{\infty}[\frac1{x+1}][/math]

 

as:

 

[math]\sum_{x=1}^3 [\frac1x] + \sum_{x=4}^{\infty}[\frac1x][/math],

 

is really only a more pedantic way of doing what Krim criticizes. It shouldn't be a problem for a calculus student; if it is, they shouldn't find it in the text book IMHO. If none of their mates can get it, they can always ask the teacher.

 

What? :confused: I don't understand you. What do you mean?

:doh: Hurried posting! :hihi:

 

Yet, by algebra and with a cancellation of exactly

equal

sums just as in your method, ...

 

If that wasn't the trouble, try multplying the series I gave by 2 (or the second one by -2) and using the distributive property. What do you get? Subtract it from the given series and contemplate the algebra...

 

In my hurry I had also left out:

Of course that difference will go to 0 as x goes to infinity. I would have appreciated an answer that showed this rather than trying to claim that the two series cancel each other out even though for any given value of the dummy variable they do not. Is this actually how most people think infinity works?

That difference going to 0, as x goes to infinity, isn't enough to say two series have the same sum, those differences might sum up to a non-zero value or even diverge. Asymptotic comparison can be useful for arguing on whether or not they converge, so long as the difference does.

 

Perhaps your book is not explaining very well, I can't tell from here, but if so I can understand your discomfort but the best way out is to base these things properly on topology.

[Pyrotex]

I vote for Craig's 11/18!!!!! :)

Me, too.

I mean that algebra isn't topology....but the best way out is to base these things properly on topology.

I have taken topology, but you're not using the term in any way I recognize. Could you spell out what you mean by "topology"? Thanks.

[Qfwfq]

A collection T of subsets of a set X is said to be a topology on X if the subsets in T satisfy the following requisite properties:

 

1. X itself and the empty set belong to T.

 

2. If a finite collection of sets belong to T, then so does their intersection.

 

3. If two or more sets belong to T, then so does their union.

 

The notion of a neighbourhood, on which the definition of limit is based, is topological. It's an abstraction of what can be, and usually is, induced by a metric.

[Pyrotex]

A collection T of subsets of a set X is said to be a topology on X if the subsets in T satisfy the following requisite properties:

 

1. X itself and the empty set belong to T.

 

2. If a finite collection of sets belong to T, then so does their intersection.

 

3. If two or more sets belong to T, then so does their union.

 

The notion of a neighbourhood, on which the definition of limit is based, is topological. It's an abstraction of what can be, and usually is, induced by a metric.

Thanks! I not only understood that, but I also meta-understood that! :hyper:

Thanks for the epiphany. Most appreciated.

Pyro

 

"It's a wonderful day in the neighborhood..." :computerkeys:

[Kriminal99]

Because the denominator's dominant term is [math]\norm x^2[/math] and not [math]\norm x^1[/math].

 

How is that signifigant other than that it allows you to avoid the logical contradiction? It seems to defend against such a problem you need justification other than just a "if you said that then other problems would occur".

 

My complaint against such a use of infinity is that it seems to contradict the algorithm we use to create infinity from our finite experiences. Why would you be able to grow one infinity slower than another just because it makes you able to solve a problem easier? This is not how we treat infinity when we want to know the limit of something like 2*(x)/(x+5) x -> inf

[Kriminal99]

The book is correct. Just because an infinite series goes to zero does not mean that the sum must converge. Summing up 1/N long enough in this series does indeed surpass any arbitrary value. One might say, that 1/N doesn't approach zero fast enoughAgain, the mere fact that that the terms get smaller and smaller is insufficient to determine whether or not the infinite series sum will approach some particular finite value, or approach infinity. The rate at which the terms decrease is very important.

 

Ok so the first one has a sum that surpasses any arbitrary value (with a given precision) and the second one does not. That is very interesting. This is some function of our number system isn't it? IE if you are only concerned with 5 decimal places, and the limit of the second one is .61111, then it is in fact still increasing, but no matter how much it increases that 5th decimal will never go to 2... Am I right in guessing that it is not only that the 5th or 6th decimal place will never change, but also that any given decimal place will eventually become fixed for any convergent series?

 

This seems like an arbitrary way to determine convergence or divergence. It would change depending on what number system you were using...

[Pyrotex]

Ok so the first one has a sum that surpasses any arbitrary value (with a given precision) and the second one does not. That is very interesting. This is some function of our number system isn't it?

No. All the equations we're using, indeed, in all of algebra and calculus, are independent of number system. They work the same whether you are using base 10, base 2, base 16, whatever.

IE if you are only concerned with 5 decimal places, and the limit of the second one is .61111, then it is in fact still increasing, but no matter how much it increases that 5th decimal will never go to 2... Am I right in guessing that it is not only that the 5th or 6th decimal place will never change, but also that any given decimal place will eventually become fixed for any convergent series?

I didn't make myself clear. When I said "a sum that surpasses any arbitrary value" I was not talking about precision. I was saying the sum goes to infinity. For example, you give an arbitrary value of 10, the sum eventually exceeds that; you give 10,000, the sum eventually exceeds that; you give 10^10,000, the sum eventually exceeds that, and so on.

[Kriminal99]

No. All the equations we're using, indeed, in all of algebra and calculus, are independent of number system. They work the same whether you are using base 10, base 2, base 16, whatever.I didn't make myself clear. When I said "a sum that surpasses any arbitrary value" I was not talking about precision. I was saying the sum goes to infinity. For example, you give an arbitrary value of 10, the sum eventually exceeds that; you give 10,000, the sum eventually exceeds that; you give 10^10,000, the sum eventually exceeds that, and so on.

 

I am afraid you are the one who misunderstood. The divergent sum surpasses those values. But the convergent one is ever increasing - it just doesn't increase fast enough such that the reduction in increase causes the increase to be delegated to ever more decimal places.

 

You should be able to calculate the threshold exactly, based on the number system. It might be convenient if it was not dependent on the number system, but here that does not seem to be the case. Which is why I was calling it arbitrary and also why I would have difficulty believing that it is the popular criteria for convergent/divergent.

 

Obviously any reduction of 1/10th the previous increase would be a convergent sum... I wonder if this is the threshold.

[CraigD]

But the convergent one is ever increasing - it just doesn't increase fast enough such that the reduction in increase causes the increase to be delegated to ever more decimal places.

This is a good intuitive observation – if the most significant digit of the differences between consecutive terms appears to be moving to the right faster than in the difference in a list of partial sums, which is visually very striking, a series based on a simple formula usually converges.

You should be able to calculate the threshold exactly, based on the number system.

No. As Pyrotex noted,

All the equations we're using, indeed, in all of algebra and calculus, are independent of number system. They work the same whether you are using base 10, base 2, base 16, whatever.

A couple important detailed points should be made about this statement.

 

A scheme for representing a number in visual form, such as “base 2” “base 10” “base 16” or “base 73451” is a numeral system. Examples of number systems are the natural numbers, the integers, the rationals, the constructables, the reals, the complex numbers, etc.

 

“All of algebra and calculus” involves the “ordinary” arithmetic operations, such as addition, multiplication, exponentiation, and logarithms. It is possible to define mathematical functions and operations that are dependent on the numeral system (eg: “the Nth digit of X”) but such functions are not usually considered arithmetic operations, but rather operations that can be constructed using arithmetic operations (eg: “the Nth digit of X” [math]= (X / b^{N-1}) \bmod b[/math] where “[math]/[/math]” is integer division, and [math]b[/math] is the numeral system’s base (eg: 2, 10, 16, 73451), so really these are just ordinary expressions that explicitly contain a reference to the numeral system used.

 

It might be helpful in understanding this to note that a number represented in any numeral system can, by definition (a numeral system must unambiguously represent a number) be converted to any other numeral system, making the choice of numeral system irrelevant.

You should be able to calculate the threshold exactly …

There is, however, a simple technique for determining if many kinds of series converges – compare it to the harmonic series ([math]\frac11 +\frac12 +\frac13 +\frac14 + . . .[/math]), or, more generally, determine a “p-series” ([math]\frac1{1^p} +\frac1{2^p} +\frac1{3^p} +\frac1{4^p} + . . .[/math]) for it. For all [math]p \gt 1[/math], the series converges. For all [math]p \le 1[/math], it diverges.

 

Note also that any multiple of a convergent or divergent series remains convergent or divergent. [math]\frac1{100} +\frac1{200} +\frac1{300} +\frac1{400} + . . .[/math] still diverges to infinity, while [math]\frac{100}{1} +\frac{100}{2^{1.01}} +\frac{100}{3^{1.01}} +\frac{100}{4^{1.01}} + . . .[/math] still converges to a finite number.

[Kriminal99]

I think it would be much easier to just say that a series diverges if the ratio between subsequent terms approaches 1 as you get farther and farther out. But that is not what I was talking about.