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Old 2005-10-22, 21:54   #1
jinydu
 
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Default 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
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Old 2005-10-22, 23:52   #2
Citrix
 
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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
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Old 2005-10-23, 01:26   #3
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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
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Old 2005-10-23, 02:33   #4
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What about s(n)+1?

Hint:- look for regions where 2 and 3 are not factors of s(n) or s(n)+1
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Old 2005-10-24, 07:27   #5
mfgoode
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Question 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
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Old 2005-10-24, 10:16   #6
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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
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Old 2005-10-24, 10:30   #7
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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
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Old 2005-10-24, 12:16   #8
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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.
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.

Last fiddled with by Wacky on 2005-10-24 at 12:18
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Old 2005-10-24, 18:14   #9
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I got your point.
Thank you very much.
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Old 2005-10-24, 18:23   #10
mfgoode
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Lightbulb 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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