Find the sum of the series 1+(1/2)+(1/3)+(1/4)+(1/6)+(1/8)+(1/9)+(1/(12))+∙∙∙ where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s


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Question Number 86762 by Tony Lin last updated on 31/Mar/20

$F i n d t h e s u m o f t h e s e r i e s$ $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \cdot \cdot \cdot$ $w h e r e t h e t e r m s a r e t h e r e c i p r o c a l s$ $o f t h e p o s i t i v e i n t e g e r s w h o s e o n l y$ $p r i m e f a c t o r s a r e 2 s a n d 3 s$
Commented by Prithwish Sen 1 last updated on 31/Mar/20

$(1 + \frac{1}{2} + \frac{1}{2^{2}} + . . . . . . .) (1 + \frac{1}{3} + \frac{1}{3^{2}} + . . . . .)$ $= \frac{1}{1 - \frac{1}{2}} x \frac{1}{1 - \frac{1}{3}} = 3 please check .$
Commented by redmiiuser last updated on 31/Mar/20

$3$
Commented by Tony Lin last updated on 31/Mar/20

$\frac{1}{1 - \frac{1}{2}} \times \frac{1}{1 - \frac{1}{3}} = 2 \times \frac{3}{2} = 3$
Commented by Prithwish Sen 1 last updated on 31/Mar/20

$sorry i fix it .$
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