Evaluate the infinite geometric series: \(\frac{3}{2}-\frac{2}{3}+\frac{8}{27}-\frac{32}{243}+\dots\)

ant101 Dec 27, 2018

#1**+2 **

Notice that we can split this up as

3/2 + 8/27 +....... and

-2/3 - 32/243 + ....... = - ( 2/3 + 32/243 + .....)

The geometric ratio of the first series is (2/3)^4 = 16/81

So.....the sum of this series = (3/2) / ( 1 - 16/81) = (3/2) / ( 65/81) = 243/130

The geometric ratio of the terms in the parentheses in the second series is ( 2/3)^4 = 16/81

So the sum of the second series is

- [ (2/3) / ( 1 - 16/81 ) ] = - ( 2/3) / ( 65/81) = - (2/3) (81/65) = - 54 /65 = -108 / 130

So....the sum is [243 - 108] / 130 = 135 / 130 = 27 /26

CPhill Dec 27, 2018

#2**+2 **

Best Answer

We can also use our handy formula: \(\frac{n}{1-r}.\) Since the series has a first term of \(\frac{3}{2}\) and a common ratio of \(\frac{-4}{9}\), we attain \(\cfrac{\frac{3}{2}}{1-\left(\frac{-4}{9}\right)}=\boxed{\frac{27}{26}}.\)

tertre Dec 27, 2018