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-

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}} + \dots \dots .) (1 + \frac{1}{3} + \frac{1}{3^{2}} + \dots . .)

= \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 .