Question

how do I find out whether an infinite geometric series converges and how do I find its sum. for example:6+3+3/2+3/4+...

Answer 1

We are given the following infinite geometric series

`6+3+(3)/(2)+(3)/(4)+\cdots`

The general form of an infinite geometric series is given by

`a_1+a_1r+a_1r^2+a_1r^3+\cdots`

Where a1 is the first term and r is the common ratio.

The first term is equal to 6

The common ratio is the division of any two consecutive numbers in the infinite geometric series.

`(3)/(6)=(1)/(2)`

You can also take any other two consecutive numbers in the series and you will get the same common ratio.

`((3)/(2))/(3)=(3)/(2)\cdot(1)/(3)=(3)/(6)=(1)/(2)`

As expected, we still got the same common ratio as before.

How do I find out whether an infinite geometric series converges?

An infinite geometric series converges if the absolute value of the common ratio is less than 1.

`\begin{gathered} |r|<1 \\ (1)/(2)<1 \\ 0.5<1 \end{gathered}`

Since the absolute value of the common ratio is less than 1, the infinite geometric series converges.

How do I find its sum?

The sum of this infinite geometric series is given by

`S=(a_1)/(1-r)`

Substitute the values of first term a1 and common ratio r.

`S=(6)/(1-(1)/(2))=(6)/((1)/(2))=6\cdot(2)/(1)=12`

Therefore, the sum of this infinite geometric series is 12.