# Wikipedia:Reference desk/Archives/Mathematics/2014 April 6

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# April 6

## sum of reciprocals of numbers with k different prime numbers diverges for all k?

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Is it true that for all k, the sum of the reciprocals of the numbers which contain k factors all different diverges? So for example for k = 4, the sequence starts out 1/(2*3*5*7) + 1/(2*3*5*11) + 1/(2*3*5*13) + 1/(2*3*7*11) +... Naraht (talk) 14:20, 6 April 2014 (UTC)

It diverges. Use induction on k. Sławomir Biały (talk) 14:43, 6 April 2014 (UTC)
Why on k? By the time we get to large 4 factor numbers, we will be adding 1/(2*3*5*553105253) which is quite small. It isn't obvious to me. -- SGBailey (talk) 19:14, 6 April 2014 (UTC)
Well, so the first thing you have to know is that Σ(1/p) diverges, albeit quite slowly (IIRC the sum of the first n reciprocal primes is asymptotic to a constant times log(log(n)). From there it should be pretty easy. --Trovatore (talk) 19:42, 6 April 2014 (UTC)
I am not going by the induction argument, but just as a nonrigorous intuition on SGBailey's statement: given that the sum of the reciprocals of the primes diverges, and the sum without the constraint that the k factors be different is the kth power of this, and hence this diverges. Those denominators that have duplicated factors forms a vanishingly small part of the series, so removing them to get the original sum should not change the divergence. While this is hand-wavy, it should supply an adequate countering intuition. —Quondum 19:55, 6 April 2014 (UTC)
You're making it waaayy too hard. It's trivial; you're missing an obvious point. --Trovatore (talk) 20:17, 6 April 2014 (UTC)
Which is why you're giving no hints. Yes, I can see a simple proof for all k, but it does not use induction. —Quondum 20:59, 6 April 2014 (UTC)
Alright, let's go ahead and give the hints, or actually the whole answer, now that I've had my coffee and French toast. The point is that, taking the k=4 case as an example, the series contains a subseries that looks like 1/(2*3*5) times the series (1/7+1/11+1/13+1/17+...). The latter series diverges because it's just the sum of the reciprocal primes, leaving off the first three terms. And if you multiply a divergent sequence by a nonzero constant, it's still divergent. I agree, it doesn't "use induction", at least at the level of natural-language proof (it probably does if you try to formulate it as a formal derivation in Peano arithmetic, but then so does practically everything, so that's not very interesting). --Trovatore (talk) 21:24, 6 April 2014 (UTC)
Thank you, that makes sense. I was assuming that something inductive was needed, but the subseries as a constant times a subset of the primes minus k-1 terms it much more direct.Naraht (talk) 22:28, 6 April 2014 (UTC)

## Small circles on a 2-sphere

I am not a mathematician. I am involved in some applied aspects due to my hobby so to speak although it is more serious than a simple hobby. Also I want to reassure you, it is not a homework. I read the article "Sphere" in the Wikipedia and could not find what I need.

Imagine a 2-sphere of radius R = 1 with a North Pole and a great circle which is the Equator. Given a point on the sphere which is distinct from the North Pole and the Equator (it is defined by an angle θ between a radius pointing to the point in question and the one going to the North Pole) I want to know the length (in relative units) of a small circle running through this point parallel to the equator. I would appreciate if someone would give me this answer. Thanks, - --AboutFace 22 (talk) 15:01, 6 April 2014 (UTC)

Well, the length of the equator is 2piR, so just 2pi in your R = 1 case. To reduce that by the angle from the north pole, you'd need to put a sine in there. I'm thinking it would be squared, so sin2(θ)2pi. Can somebody else verify this ? (I believe this works for circles in the southern hemisphere, too.) StuRat (talk) 15:11, 6 April 2014 (UTC)
No, it's $2 \pi R \sin \theta$. Looie496 (talk) 15:14, 6 April 2014 (UTC)
Yes, I just grabbed a globe and verified that. I measured 29 inches at the equator and 20.5 inches at the 45 degree mark, giving me a ratio of 0.7069, right on with the sin(45) = 0.7071. StuRat (talk) 15:19, 6 April 2014 (UTC)

Wow! So fast! Thank you very much. StuRat is here as always! I don't see the proof though. It seems to be more like intuitive? I hope you guys are correct. It is definitely correct in two extreme cases, theta = 0 and theta = 90. Looie496, thanks. Thanks again. --AboutFace 22 (talk) 15:24, 6 April 2014 (UTC)

The proof is trivial, although our latitude article is a bit technical to serve as a good starting point. See instead (for instance) this diagram; if you complete the (right) triangle formed by the radius to a point on the surface, a line through that point parallel to the equatorial plane, and a portion of the axis, you will see that the distance from the surface point to the axis is $R\sin\theta$ (your θ is the colatitude that mathematicians typically use). That distance is the radius of the parallel at that latitude, so the circumference is $2\pi R\sin\theta$, as given. --Tardis (talk) 16:29, 6 April 2014 (UTC)
Oh, and note that this is only for a theoretically perfect sphere. The Earth is actually a lumpy oblate spheroid, so your mileage may vary. StuRat (talk) 17:09, 6 April 2014 (UTC)

Again, I appreciate it. Thanks, --AboutFace 22 (talk) 19:58, 6 April 2014 (UTC)

Well, I am back here. Either there is something wrong with the formula or I am making a mistake somewhere. First I have a small globe in front of me. Then I decided to calculate the latitude as you call it or a small circle on the sphere which is just 3 degrees removed from the North Pole. It is a tiny distance, in fact. This is what I get. The sine ( θ ) = 0.052359... I don't know how to use calculators, it is easy for me to write a short piece of code, so it is C#. Then using your formula I multiply 6.28 * 0.052359 and I get 0.3284. It seems too large a value for me. Could it be true? Thanks, --AboutFace 22 (talk) 21:29, 6 April 2014 (UTC)

Yup, correct (you missed a digit) for the circumference (what I assume you mean by "length") of the circle. This is 5.2% of the circumference of the equator. Colatitude: 3°, Latitude: 87°, circumference of circle of latitude: 0.3288365⋅R, where you took R = 1. —Quondum 21:53, 6 April 2014 (UTC)

Quondum, thank you but I do not understand what you are trying to say? First you said "correct." In what sense am I correct? That I made a mistake or the result I posted is correct? And what digit did I miss? Could you be more specific? Yes, I am talking about circumference. It is the length of the small circle I meant. On the other hand my result appears to be around 5% of the Equator (great circle). --AboutFace 22 (talk) 22:05, 6 April 2014 (UTC)

The formula looks right to me. To convince yourself that everything is OK, try what I did: Get a large globe, run a string along it at the equator, and measure the length of the string. Then repeat this while making a circle 3 degrees from the N pole (it helps if your globe has latitude markings). The ratio of the two lengths should be the sine of 3 degrees.
Note that a larger globe and larger angle on the globe from the pole will both reduce the relative error. StuRat (talk) 22:19, 6 April 2014 (UTC)
The value 0.32884 is the correct value for the circumference of a circle around the north pole at colatitude 3° on a sphere of radius 1. Your value differed from this in that it omitted one of the '8' digits. Perhaps the little circle seems less that 5% of the size of the equator because the area of the discs differs so much (area ratio 1:365). —Quondum 22:22, 6 April 2014 (UTC)

Thank you StuRat and especially Quondum. Now it is all clear. A rephrased statement did the trick. --AboutFace 22 (talk) 00:41, 7 April 2014 (UTC)

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