Find the sum of the series 1 12 13 14 16 18 19 112 Wh

Find the sum of the series 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12 +...Where the denominators are positive integers whose only prime factors are 2s and 3s. Find the sum of the series 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12 + 1/5 +...This time the denominators are positive integers whose only prime factors are 2s, 3s and 5s.

Solution

1.  change it into two series

one being the powers of two and the other powers of three

1+1/2+1/4+1/8+.... converges to a/(1-r) = 1 / (1- 1/2) = 2

1+ 1/3 + 1/9 + 1/27+....coverges to 1 / (1-1/3) = 3/2

So sum is 2 + 3/2 = 7/2

, if you have all of them, then you will also have the sum of
1/2 + 1/2*3 + 1/2*3^2 + . . . .converges to 1/2 / (1-1/3) = 3/4

1/4 + 1/4*3 + 1/4*3^2+ . . .converges to 1/4 (1-1/3) = 3/8

add sums up: 3/2 + 3/4 + 3/8 + 3/16 + ...coverges to 3/2 / (1 - 1/2) = 3

SUM is themselves a series, therefore sum = 3

2.

change it into three series

one being the powers of two and the other powers of three

1+1/2+1/4+1/8+.... converges to a/(1-r) = 1 / (1- 1/2) = 2

1+ 1/3 + 1/9 + 1/27+....coverges to 1 / (1-1/3) = 3/2

1+ 1/5 + 1/25 + 1/125+....coverges to 1 / (1-1/5) = 5/4

So sum is 2 + 3/2+ 5/4= 19/4

, if you have all of them, then you will also have the sum of
1/2 + 1/2*3 + 1/2*3^2 + . . . .converges to 1/2 / (1-1/3) = 3/4

1/4 + 1/4*3 + 1/4*3^2+ . . .converges to 1/4 (1-1/3) = 3/8

1/2 + 1/2*5 + 1/2*5^2 + . . . .converges to 1/2 / (1-1/5) = 5/8

1/3 + 1/3*5 + 1/3*5^2 + . . . .converges to 1/3 / (1-1/3) = 5/16

add sums up: 3/4 + 3/8 + 5/8 + 5/16 + ...coverges to 3/2 / (1 - 1/2)+ 5/8 /(1-1/2)
=3+ 5/4= 17/4