Question

How do you find the sum of the infinite geometric series (3/2)-(5/6)+(7/18)-...?

Answer 1

`15/16`

Explanation:

This is not a geometric series. It has no common ratio, but the intention seems to be that the general term is:

`a_n = (-1)^(n-1)((2n+1))/((2*3^(n-1)))`

If we sum pairs of terms, then the resulting series has all positive terms, with the following expression for the general term:

`b_n = ((4n-2))/((3*9^(n-1)))`

The numerators are in arithmetic progression with initial term `2` and common difference `4`.

We can split this series into a series of series, hence into a product of a geometric series and a geometric series with constant offset.

Note first that `sum_(k=0)^oo 1/9^k = 9/8`

So we find:

`3/2-5/6+7/18-9/54+11/162-13/486+15/1458-17/4374+...`

`=2/3+6/27+10/243+14/2187+...`

`=1/3((2+2/9+2/9^2+2/9^3+...)+(4/9+4/9^2+4/9^3+...)+(4/9^2+4/9^3+4/9^4+...)+(4/9^3+4/9^4+4/9^5+...)+...)`

`=1/3(1+1/9+1/9^2+1/9^3+...)(2+4/9+4/9^2+4/9^3+4/9^4+...)`

`=4/3(sum_(k=0)^oo 1/9^k)(-1/2+sum_(k=0)^oo 1/9^k)`

`=4/3*9/8*(9-4)/8`

`=4/3*9/8*5/8`

`=15/16`