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

1 Answer
Jun 28, 2016

15/161516

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)))an=(1)n1(2n+1)(23n1)

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)))bn=(4n2)(39n1)

The numerators are in arithmetic progression with initial term 22 and common difference 44.

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 = 9/8k=019k=98

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