can someone check my work? ** series

[Guest]

Find the sum of the series where the terms are reciprocals of the positive integers whose prime factors are 2's and 3's

1+1/2+1/3+1/4+1/6+1/8+1/9+1/12+1/16+....+

I believe, the sum of the series should form a double sum of 1/2^a & 1/3^b which equals axb

sigma 1/2^a [a, 1, infinity] (Sigma 1/3^b [b, 1, infinity])=

sigma 1/2^a [a, 1, infinity] (1/1-1/3) =

3/2 sigma 1/2^a [a, 1, infinity] =

3/2 (1/1-1/2) = 3


I hope everyone can make that out - you're help is greatly appreciated!

Thanks!

 

[galactus]

If a and b are 1 to infinity, you should get rid of the 1 in your summation.

1/2+1/3+1/4+1/6+..................=2.

0 to infinity would be 3.

 

[skeeter]

the terms are reciprocals of the positive integers whose prime factors are 2's and 3's

this is a poorly worded question.

... is it 2's and 3's or is it 2's or 3's ???

2's or 3's (no other prime factors) ...

1/2 + 1/4 + 1/8 + 1/16 + ... = 1

1/3 + 1/9 + 1/27 + 1/81 + ... = 1/2

sum 1/(2^n) + sum 1/(3^n) = 3/2


2's and 3's (no other prime factors?) ...

(1/6 + 1/12 + 1/18 + 1/24 + 1/30 + ...) = sum 1/(6n) ... diverges.

 

[pka]

the terms are reciprocals of the positive integers whose prime factors are 2's and 3's this is a poorly worded question.
(1/6 + 1/12 + 1/18 + 1/24 + 1/30 + ...) = sum 1/(6n) ... diverges.

Actually Skeeter this is a rather well known series and Galactus is correct in his post.
\(\displaystyle \L
\left( {\sum\limits_{k = 0}^\infty {a_k } } \right)\left( {\sum\limits_{k = 0}^\infty {b_k } } \right) = \left( {\sum\limits_{k = 0}^\infty {c_k } } \right)\quad \mbox{where} \quad c_n = \sum\limits_{k = 0}^n {a_k b_{n - k} }\)

So finding \(\displaystyle \L
\left( {\sum\limits_{k = 0}^\infty {\frac{1}{{2^k }}} } \right)\left( {\sum\limits_{k = 0}^\infty {\frac{1}{{3^k }}} } \right) - 1\) gives sum of all reciprocals of the positive integers whose prime factors are 2's and 3's.

 

[Guest]

galactus, thank you for checking my work. i did make a mistake while typing out my problem it is suppose to be 0 to infinity.

 

[skeeter]

Actually Skeeter this is a rather well known series and Galactus is correct in his post.
\(\displaystyle \L
\left( {\sum\limits_{k = 0}^\infty {a_k } } \right)\left( {\sum\limits_{k = 0}^\infty {b_k } } \right) = \left( {\sum\limits_{k = 0}^\infty {c_k } } \right)\quad \mbox{where} \quad c_n = \sum\limits_{k = 0}^n {a_k b_{n - k} }\)

So finding \(\displaystyle \L
\left( {\sum\limits_{k = 0}^\infty {\frac{1}{{2^k }}} } \right)\left( {\sum\limits_{k = 0}^\infty {\frac{1}{{3^k }}} } \right) - 1\) gives sum of all reciprocals of the positive integers whose prime factors are 2's and 3's.

No doubt, I believe you ... I'll have to work it out for myself to "see" it.
Thanks for introducing me to this "well-known" series that was previously "unknown" by me.