- #1
DivGradCurl
- 372
- 0
Find the sum of the series
[tex] 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \dotsb [/tex]
where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s.
Well, here is my guess:
[tex] 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} [/tex]
As you can readily observe, I'm really stuck. Maybe someone could give me a tip. Any help is highly appreciated.
Thanks
[tex] 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \dotsb [/tex]
where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s.
Well, here is my guess:
[tex] 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} [/tex]
As you can readily observe, I'm really stuck. Maybe someone could give me a tip. Any help is highly appreciated.
Thanks