20051022, 21:54  #1 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Infinite Sum Problem
Here's an interesting problem from the "Problems Plus" section of my calculus textbook:
Find the exact value of the infinite sum of the reciprocals of all integers greater than 1 whose only prime factors are 2 and 3. The first few terms are: Last fiddled with by jinydu on 20051022 at 21:55 
20051022, 23:52  #2 
Jun 2003
3053_{8} Posts 
x=sum of (1/2+1/4+1/8+...)
y= sum of (1/3+1/9+1/27+...) then sum of your series is x+x*y+y Last fiddled with by Citrix on 20051023 at 00:06 
20051023, 01:26  #3 
Jun 2003
1,579 Posts 
A related problem
let s=2+3+4+6+..., sum of the series. then s(n) is prime for what n? Are there infinite primes of the form s(n)? Citrix 
20051023, 02:33  #4 
Jun 2003
11000101011_{2} Posts 
What about s(n)+1?
Hint: look for regions where 2 and 3 are not factors of s(n) or s(n)+1 
20051024, 07:27  #5  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Infinite Sum Problem
Quote:
Kindly give me the numerical value of the answer of the infinite sum then? Thanks 

20051024, 10:16  #6  
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
Quote:
Using Citrix's formulation x=sum of (1/2+1/4+1/8+...) Therefore 2*x = 1+x ==> x=1 and y= sum of (1/3+1/9+1/27+...) Therefore 3*y = 1+y ==> y= 1/2 then sum of your series is x+x*y+y = 1 + 1 * 1/2 + 1/2 = 2 

20051024, 10:30  #7  
Sep 2002
Oeiras, Portugal
1,451 Posts 
Quote:
Some terms are not divisible by BOTH 2 and 3. So you mean "integers whose only prime factors are 2 and/or 3"? If that is the case, the result shall be: x=sum of (1/2+1/4+1/8+...) = 1. y=sum of (1/3+1/9+1/27+...) =1/2. z=sum of (1/6+1/12+1/18+...) = sum of 1/6n > divergent S= x+y+z = infinity And your series should be divergent as well. Unless I missed something. Last fiddled with by lycorn on 20051024 at 10:31 

20051024, 12:16  #8  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:
However, since he explicitly stated some terms of the series, I think that we must conclude that he intended to include the above terms and use the definition "whose prime factors are not larger than 3" Quote:
The series that you describe continues (… + 1/24 + 1/30 + …) 1/30 does not fit the required conditions because 5 is a prime factor of 30. I believe that Citrix's formulation is correct. Each denominator is of the for 2^i * 3^j for some i,j >= 0, but excluding the term where (i,j = 0,0) So we get Sum(Series) = Sum(i=0,inf){Sum(j=0,inf){2^(i)*3^(j)}} 1 But this is easily transformed to (1+x)(1+y)  1 And the rest follows immediately. Last fiddled with by Wacky on 20051024 at 12:18 

20051024, 18:14  #9 
Sep 2002
Oeiras, Portugal
10110101011_{2} Posts 
I got your point.
Thank you very much. 
20051024, 18:23  #10  
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts 
Infinte Sum Problem
Quote:
Mally 

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