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 2005-10-22, 21:54 #1 jinydu     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: $\Large{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\frac{1}{16}+...}$ Last fiddled with by jinydu on 2005-10-22 at 21:55
 2005-10-22, 23:52 #2 Citrix     Jun 2003 30538 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 2005-10-23 at 00:06
 2005-10-23, 01:26 #3 Citrix     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
 2005-10-23, 02:33 #4 Citrix     Jun 2003 110001010112 Posts What about s(n)+1? Hint:- look for regions where 2 and 3 are not factors of s(n) or s(n)+1
2005-10-24, 07:27   #5
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
Infinite Sum Problem

Quote:
 Originally Posted by Citrix 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

Kindly give me the numerical value of the answer of the infinite sum then?
Thanks

2005-10-24, 10:16   #6
Wacky

Jun 2003
The Texas Hill Country

32×112 Posts

Quote:
 Originally Posted by mfgoode Kindly give me the numerical value of the answer of the infinite sum then?
TWO

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

2005-10-24, 10:30   #7
lycorn

Sep 2002
Oeiras, Portugal

1,451 Posts

Quote:
 Originally Posted by jinydu 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: $\Large{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\frac{1}{16}+...}$
There´s something unclear to me.
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 2005-10-24 at 10:31

2005-10-24, 12:16   #8
Wacky

Jun 2003
The Texas Hill Country

32·112 Posts

Quote:
 Originally Posted by lycorn Some terms are not divisible by BOTH 2 and 3. So you mean "integers whose only prime factors are 2 and/or 3"?
I think that you are correct. As stated, the terms of the form 2^(-n) and the terms of the form 3^(-n) would be omitted.
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:
 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.
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 2005-10-24 at 12:18

 2005-10-24, 18:14 #9 lycorn     Sep 2002 Oeiras, Portugal 101101010112 Posts I got your point. Thank you very much.
2005-10-24, 18:23   #10
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

1000000001002 Posts
Infinte Sum Problem

Quote:
 Originally Posted by Wacky Your "z" term is incorrect. 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.
Brilliant Wacky and excellent deduction The term where (i,J =0,0) excluded is the crux of the matter!. Keep up the good work.
Mally

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