Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I was thinking about the infinite series that goes 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ..., where the nth term is one divided by the nth prime.

I'm wondering if anyone knows (or can work out) whether it converges or diverges. And if it converges, roughly to what?

I know that the series of 1/x diverges, and that the series of 1/x² converges, and 1/P seems to be somewhere between those two.

I started comparing it with other series to see if I could get some insight, and in doing so I made up a proposal that seems true but I have no idea whether it is.

Basically, an infinite series made up of terms of 1/f(x) will always converge if you can find an integer k such that f(k+1) - f(k) is greater than any given N.

It's easy to see how this is true for x² and false for x, but for the primes it doesn't help as much.

So all in all, I've had lots of thoughts but none of them really help.

Can anyone shed light on this?

Why did the vector cross the road?

It wanted to be normal.

Offline

**Identity****Member**- Registered: 2007-04-18
- Posts: 934

Based on what you said about f(k+1) - f(k), wouldn't it converge since the distance between between primes increases as k increases? But that's just my intuition speaking there, perhaps primes don't thin out sufficiently at infinity. I'd be interested in what it converges to though...

Offline

**bossk171****Member**- Registered: 2007-07-16
- Posts: 305

I think I read on wikipeida that it diverges, not that they're a respectable source...

EDIT: It does in fact say that, I still don't know how to prove it.

*Last edited by bossk171 (2007-09-16 04:09:23)*

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

Offline

**bossk171****Member**- Registered: 2007-07-16
- Posts: 305

http://www.maa.org/editorial/euler/How% … primes.pdf

http://mathworld.wolfram.com/HarmonicSe … rimes.html

confirms the wiki article but doesn't really say why.

http://www.everything2.com/index.pl?node_id=1537535

gives a proof I don't understand.

*Last edited by bossk171 (2007-09-16 04:58:27)*

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

Offline

**Identity****Member**- Registered: 2007-04-18
- Posts: 934

And silly me was trying to estimate its convergence. Until I read bossk171's post I thought it would converge to *e*.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

This proof is a bit easier to understand.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**bossk171****Member**- Registered: 2007-07-16
- Posts: 305

I only sorta get that.

As long as we're on the topic of infinate series, I played with the series:

1/1 + 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/ 13 + 1/21 +...

The reciprocals of Fibonacci numbers. I brute forced it in excel and it definately seemed to converge...

EDIT: I spoke too soon, the wikipedia article says that it does converge but no one knows how to express the answer as anything but an estimate of its value: 3.35988...

*Last edited by bossk171 (2007-09-16 07:20:32)*

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I only sorta get that.

Which statement do you not get?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**bossk171****Member**- Registered: 2007-07-16
- Posts: 305

"Well, there's an elementary theorem of calculus that a product (1-a1)...(1-ak)... with ak->0 converges to a nonzero value iff the sum a1+...+ak+... converges"

I have taken calc, but I don't remember that, or have never seen it. If I could get past that, the rest pretty much makes sense.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Ah, then you do follow this proof, it's just a proof for theorem used in it which you don't. I too, have yet to see proof of that theorem. Anyone care to try or dig one up?

Edited to add: And by "calc" I believe they were referring to analysis, as I don't think that statement would be very useful in any introductory calculus class.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**bossk171****Member**- Registered: 2007-07-16
- Posts: 305

Some standard work shows we can open the brackets

I'm not sure if I've hear the phrase "open the brackets" before. Does this mean something common and it's just the slang I'm stuck on, or is this something I haven't learned yet?

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

"Open the brackets" simply means reorder the terms. So long as you are summing all positive terms, any permutation (reordering) will work, so long as all terms eventually appear in the new sequence. But when you come to alternating series, the same isn't true. This is why you must be careful.

Offline

**bossk171****Member**- Registered: 2007-07-16
- Posts: 305

Yeah, thanks! it all makes sense now. Except for that initial theorm.

*Last edited by bossk171 (2007-09-17 03:12:46)*

Offline

A nice result is that the partial sums of that series is:

or, more precisely:

where M is the so-called Meissel-Merten's constant, with value approximately equal to 0.2614972128476427837554268386086958590516...

This result is known as Merten's second theorem.

Offline

Pages: **1**