## A prime sum

Posted by Chris on March 25, 2011 – 7:21 am

If H is the infinite set of all the positive integers having, at most, prime factors 2, 3 and 5. Show that the sum of the reciprocals of the elements of H = 15/4.

For clarity,the first few elements of H are: 1,2,3,4,5,6,8,9,10,12,…

March 25th, 2011 at 7:25 am

Hi slavy. I know this’ll be a doddle for you.

March 25th, 2011 at 11:37 am

Well I can’t get it to show up correctly, but if you put this in Microsoft Word and change it to an equation (professional) it should show up correctly.

∑_(a=1)^∞▒∑_(b=1)^∞▒∑_(c=1)^∞▒1/((a*2)(b*3)(c*5))

Basically, there are 3 sums in a row. The first is a=1->inf. The second is b=1->inf. And the third is c=1->inf.

The expression to sum is 1/[(a*2)(b*3)(c*5)].

Our job is to show that all of that equals 15/4 = 3.75

This is looking like a combination of 3 Harmonic series, which is probably why you gave us the clue of calling it H.

From quick memory, there seems to be a factor of 1/2 when solving. Something like the sum of i as n=1->inf = 1/2n(n+1). I don’t have the time to show all of my work right now, but if you factor out the common denominator of all numbers (2*3*5) and apply the (3) 1/2 factors: 2*3*5*(1/2*1/2*1/2) = 15/4. That may be the incorrect rule to allpy to this situation. I’m looking forward to the solution.

March 25th, 2011 at 11:39 am

for clarity, that was 1/2*n*(n+1)

March 25th, 2011 at 12:15 pm

Hi DP. Are you using * as a multiply or raise to the power? For raise to the power use ^.

~~Whichever, you haven’t got it right~~. But you are warm.Rather than just write down sums, show how you derived them.

What is the ▒ meant to be? A ( perhaps!

March 25th, 2011 at 2:13 pm

* is multiply. the fuzzy looking box is part of the equation builder. I think it is just where the next term is supposed to be. Had I used “E” for the sum symbol, it would look something like EEE 1/[(a*2)(b*3)(c*5)] where each E (or sum) has limits of each variable (a, b, & c) from 1 to infinity.

March 25th, 2011 at 2:18 pm

Oh…how I came up with the terms…

I know that the first term is 1/(2*3*5). The next is 1/(2*3*(2*5))…

it is difficult to see the pattern in my two examples, but the denominator is some constant times 2, times some consant times 3, times some constant times 5. Occasionally they are the same constant like the first term where they are all 1.

I represent these different constants by a, b, and c.

so the denominator becomes (a*2)(b*3)(c*5).

For the first round, let a=1, b=1, and c from 1 to inf.

second round a=1, b=2, c from 1 to inf.

repeat as b goes to inf, then again as c goes to inf.

March 25th, 2011 at 2:25 pm

I’m embarrassed to say that I’ve goofed the question. Now I understand where your expression came from. I’ve fixed it.

March 25th, 2011 at 3:09 pm

1 + 1/2 + 1/4 + … = 2

1 + 1/3 + 1/9 + … = 3/2

1 + 1/5 + 1/25 + … = 5/4

Sum of reciprocals of all numbers having factors 2, 3 and 5

= (1 + 1/2 + 1/4 + …) * (1 + 1/3 + 1/9 + …) * (1 + 1/5 + 1/25 + …)

= 2 * 3/2 * 5/4 = 15/4

March 25th, 2011 at 3:20 pm

Hi WIz. Yep . I nearly mentioned that I thought you might breeze it too. However, you haven’t explained what you did, or why your equations are right. There’s a slightly hard way and there’s a more sophisticated way of establishing their truths.

I had goofed the question. But the answer to the goofed version would have been 1/8.

March 25th, 2011 at 3:52 pm

Hi Chris, I will bi silent I am still eating this one third of a banana that we robbed the poor camel with, so my mouth is busy

March 25th, 2011 at 4:01 pm

By the way, I know I am terrible with numbers and arithmetic but is there any slight chance that the answer is 11/4, instead of 15/4?

March 25th, 2011 at 4:14 pm

Hi slavy. You’re right. Unfortunately, I hadn’t been careful, as usual , when setting the question. I’ve now modified it to retain the original answer.

In case anybody’s wondering, I hadn’t included the 1 in the set H.

March 26th, 2011 at 8:16 am

I’ve not been paying close enough attention to this one. I confused myself with Riemann’s Zeta function – there is no sophisticated way of doing this (I had the fundamental theorem of algebra in mind).

So my apologies to DP, he hadn’t made a mistake. There again, he still had a fair way to go.

Despite having introduced H, I could just as easily have said the set was the set of the reciprocals.

Wiz wrote S = (1 + 1/2 + 1/2² + …)*(1 + 1/3 + 1/3² + …)*(1 + 1/5 + 1/5² + …).

If you multiply that out (in your head will do), it is obvious that you will get the required sum – i.e. every possible combination of 2,3 and 5 and every possible power of each will be produced (once and only once).

Using the well known identity 1/(1-x) = 1 + x + x² + x³ + … (|x| < 1)

=> S = 1/(1 – 1/2)*1/(1- 1/3)*1/(1-1/5) = 1/(1/2) * 1/(2/3) * 1/(4/5) = 15/4

March 27th, 2011 at 6:32 am

the 8th comments was wrong..

cause if we take the set of prime number of 2,3 and 5 then definetly their would be a copy of (6n,15n,10n:n is any positive integer) so we have to remove that copies first then we go further for adition of 1+1\2+1\4 and all these…

so from my point of view the set that contains all numbers which are primefactor of 2,3and 5 will be

{(2n,3n,5n)-(6n,15n,10n)+(30n)} where n is from 1 to n

so the sumation of inverse of this set will be

((1/2)summation(1/n)+(1/3)summation(1/n)+(1/5)summation(1/n)-(1/6)summation(1/n)-(1/15)summation(1/n)-(1/10)summation(1/n)+(1/30)summation(1/n)) where n is from 1 to n

by simplifying, we get

((1/2)+(1/3)+(1/5)-(1/6)-(1/15)-(1/10)+(1/30))*summation(1/n)

(22/30)*summatiob(1/n)

know summaition of (1+(1/2)+(1/3)+…) will be

ln(n + 0.5) + 0.5772 + 0.03759/(n*n + 1.171)

putting n tend to infinity we get

(22/30)*(5.11363636….)=3.7499999999

and 15/4=3.75

hence we get the correct answer

March 27th, 2011 at 7:25 am

Hi vedsar. There are no copies in the product. Can you give a definite example of a copy and how it came about?

Also, Sum(n=1 to k, 1/n) → ∞ as k → ∞. i.e. Sum(1/n) is indeterminate.

March 28th, 2011 at 7:30 am

okey consider the set having 2 as prime factor

{2,4,6,8,10,12,14,16,18…..}

having 3 as prime factor

{3,6,9,12,15,18,21…}

and 5 as prime factor

{5,10,15,20,25……}

know

6n i.e

{6,12,18,24……}repeats twice, so we remove it once….

similarly

15n and 10n are also remove….

but in the mean while the set of 30n is removed twice so again once we add the set{30,60,90………}

actually its just like

(aUbUc)=a+b+c-intersection of(a,b)-intersection of(b,c)-intersection of(c,a)+intersection of(a,b,c)

March 28th, 2011 at 8:23 am

Actually Chris, I’m getting the exact opposite from vedsar. In your solution, there are several numbers left out. In your newest version of the problem, you say “the first few elements of H are: 1,2,3,4,5,6,8,9,10,12,…”

Where in your solution does the number 6 (for example) show up in the denominator? and 10?…

Does your solution completely leave out all numbers that share prime factors 2 and/or 3 and/or 5?

I do agree that even as originally stated, I was not close to the answer. I also didn’t have a plan of making a general expression for when my a, b, or c = 0. And you show being able to use those numbers, since you changed the problem to say “at most”.

March 28th, 2011 at 9:39 am

Hi vedsar. You’ve invented sets that I didn’t specify – that’s why you have copies. The general element of H, the only set I specified, is:

(2^p)(3^q)(5^r) where p,q,r = 0,1,2,3,…, independently.

Here’s a longer example:

H = {1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36,40,…}.

I’ve only omitted elements above 40.

Hi DP. the sum we want is the sum of the reciprocals of the elements of H.

The 1/6 comes from (… + 1/2 + …)(… + 1/3 + …)(1 + …), where I’ve only shown the relevant terms.

There is no overall denominator. For clarity the sum we want is:

1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/8 + 1/9 + 1/10 + 1/12 + …

My apologies for having goofed and then tweaked the question. The original wording suggested that each element had factors of 2, 3 and 5, but didn’t make it clear that e.g. 7 couldn’t be a factor – i.e. I wasn’t careful grammatically. I’d also, unintentiaonally left out the 1 in the sample.

March 28th, 2011 at 10:25 am

Ok, yep that clears it up.

So with the new info, I also get 1/(1 – 1/2) * 1/(1 – 1/3) * 1/(1 – 1/5) = 2 * 3/2 * 5/4 = 15/4 = 3.75

So no harmonic series? Sad.

March 28th, 2011 at 10:54 am

Hi DP. At least vedsar gave the harmonic series a pretty good airing.

March 28th, 2011 at 12:18 pm

Hi DP. If you extend H to include all primes, then you will get the harmonic series, and find (notionally) that:

1 +1/2 +1/3 +1/4 +1/5 +… = 1/((1 -1/2)(1 -1/3)(1 -1/5)(1 -1/7)(1 -1/11)…(1 -1/p)…), where p represents a prime. That is intimately related to a very important discovery by Euler, which culminates in the famous Riemann Zeta function, and the most important of outstanding conjectures. $1,000,000 and a place in history if you can prove the conjecture is true.

Hi vedsar. Your sets include non-admissable terms, such as 14, as that has a factor of 7.

March 29th, 2011 at 7:02 am

oh, sorry cris i forgotted that “at most” word. which neglects 14, and also many more terms….

March 29th, 2011 at 8:32 am

Hi vedsar. No problem. At least you now don’t have to force the harmonic series to become convergent

I was a bit naughty in my last post. I chose the only non-convergent generalised harmonic series. If I’d gone for Sum(n=1 to ∞, 1/(n

^{s})) = Product(all prime p,1/(1 – 1/(p^{s})), then that’s OK, as long as Re(s) > 1. This can be analytically continued for all s ≠ 1.