Sum of Series with Prime Factors 2 and 3 | Reciprocal Positive Integers

[DivGradCurl]

Find the sum of the series

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \dotsb$$

where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s.

Well, here is my guess:

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \dotsb = 1 + \sum _{n=1 } ^{\infty} \left( \frac{1}{2} \right) ^n + \sum _{n=1 } ^{\infty} \left( \frac{1}{3} \right) ^n + \sum _{n=1 } ^{\infty} \left( \frac{1}{2\cdot 3} \right) ^n + \mbox{ ? } = \sum \frac{1}{2^x \cdot 3 ^y}$$

As you can readily observe, I'm really stuck. Maybe someone could give me a tip. Any help is highly appreciated.

Thanks

 

[Gokul43201]

Look at :

$$1 + \sum _{n=1 } ^{\infty} \frac{1}{2^n} + \sum _{n=1 } ^{\infty} \frac{1}{3^n} + \left( \sum _{n=1 } ^{\infty} \frac{1}{2^n} \right) \cdot \left( \sum _{n=1 } ^{\infty} \frac{1}{3^n} \right)$$

 

[Hurkyl]

From the final sum you got, it looks like you're on the right track (though the stuff in the middle looks wrong). It's actually pretty easy to evaluate; the method might be easier to see if you didn't use shorthand notation.

 

[krab]

Let S be the sum of powers of 1/2. Then besides S you will also get S/3, S/3^2, etc. Add them all together and you get...?

 

[Gokul43201]

Let S be the sum of powers of 1/2. Then besides S you will also get S/3, S/3^2, etc. Add them all together and you get...?

This is much nicer !

 

[DivGradCurl]

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \dotsb$$

(1) Let S be the sum of powers of 1/2.

$$S = \sum _{n=1} ^{\infty} \left( \frac{1}{2} \right) ^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dotsb = \frac{1}{1-\frac{1}{2}} - 1 = 1$$

(2) Then besides S you will also get S/3, S/3^2, etc.

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \dotsb - S = 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{9} + \frac{1}{12} + \dotsb = S + \frac{S}{3} + \frac{S}{6} + \frac{S}{9} + \frac{S}{12} + \dotsb$$

(3) Add them all together and you get...?

If I followed your directions correctly, I should be able to add all the numbers. First, I sum the series where the terms are the reciprocals of the positive integers whose only prime factors are 3s:

$$S + \frac{S}{3} + \frac{S}{6} + \frac{S}{9} + \frac{S}{12} + \dotsb = S \sum _{n=0} ^{\infty} \left( \frac{1}{3} \right) ^n \underbrace{ + \frac{S}{6} + \frac{S}{12} + \dotsb} _{OUTSIDE}$$

But, as you can see, there are some terms left out of the series above. However, it seems to be implied on (2) it is the only one required to do the job ("you will also get S/3, S/3^2, etc"). According to Gokul43201, your method works fine, so I must have done quite a few mistakes up to this point. Could you please clarify that?

Thanks

 

[arildno]

You could do a follow-up on Gokul's initial method and show that your sum may be written as:
$$S=(\sum_{i=0}^{\infty}\frac{1}{2^{i}})(\sum_{j=0}^{\infty}\frac{1}{3^{j}})$$

(Note that this is just what krab implied..)

 

[DivGradCurl]

It makes sense. So, we get

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \dotsb = (\sum_{i=0}^{\infty}\frac{1}{2^{i}})(\sum_{j=0}^ {\infty}\frac{1}{3^{j}})$$

Thank you

 

[arildno]

Correct.
I see I used "S" in a different sense than krab did, but fortunately, you got what I meant.
Note that Gokul's initial result is found through:
$$(\sum_{i=0}^{\infty}\frac{1}{2^{i}})(\sum_{j=0}^{\infty}\frac{1}{2^{j}})=(1+\sum_{i=1}^{\infty}\frac{1}{2^{i}})(1+\sum_{j=1}^{\infty}\frac{1}{2^{j}})$$
and then expanding the brackets.

 

[learningphysics]

If I followed your directions correctly, I should be able to add all the numbers. First, I sum the series where the terms are the reciprocals of the positive integers whose only prime factors are 3s:

$$S + \frac{S}{3} + \frac{S}{6} + \frac{S}{9} + \frac{S}{12} + \dotsb = S \sum _{n=0} ^{\infty} \left( \frac{1}{3} \right) ^n \underbrace{ + \frac{S}{6} + \frac{S}{12} + \dotsb} _{OUTSIDE}$$

But, as you can see, there are some terms left out of the series above. However, it seems to be implied on (2) it is the only one required to do the job ("you will also get S/3, S/3^2, etc"). According to Gokul43201, your method works fine, so I must have done quite a few mistakes up to this point. Could you please clarify that?

Thanks

To get all the terms in your series, you need every power of (1/2) multiplied by every power of (1/3). This follows from the definition of your series... So one way to get all the terms is to take all the different powers of 1/2, multipling by (1/3)^0, then adding all the powers of 1/2 multiplied by (1/3)^1, then adding all the powers of 1/2 multplied by (1/3)^2 etc... In this manner you get every power of (1/2) multiplied by every power of (1/3)

So your sum is:
(1+1/2 + 1/3 + 1/4 + 1/6 + 1/8 +1/9 +1/12 + ...)
= 1*(1+1/2 + 1/4 + 1/8 + 1/16 +...) + (1/3)(1+1/2 + 1/4 + 1/8 + 1/16 +...) + (1/9)(1+1/2 + 1/4 + 1/8 + 1/16 +...) +...

S=(1+1/2+1/4+1/8+...)

Substitue in S to the original sum, so the original sum=
1*S + (1/3)*S + (1/9)*S +...