# How do you find the sum of the infinite geometric series given #-8-4-2-...#?

##### 1 Answer

#### Explanation:

Note that the given series is a geometric series with initial term

#(-4)/(-8) = (-2)/(-4) = 1/2#

We find:

#-8-4-2-... = (2-1)(-8-4-2-...)#

#color(white)(-8-4-2-...) = (-16-8-4-...) - (-8-4-2-...)#

#color(white)(-8-4-2-...) = -16#

...assuming the sum converges.

In the case of a general infinite series, we should usually look at partial sums and make sure that they converge to a limit.

The general term of a geometric series can be written:

#a_n = a * r^(n-1)#

where

Then we find:

#(1-r) sum_(n=1)^N a_r=(1-r) sum_(n=1)^N a * r^(n-1)#

#color(white)((1-r) sum_(n=1)^N a_r)=sum_(n=1)^N a * r^(n-1) - r sum_(n=1)^N a * r^(n-1)#

#color(white)((1-r) sum_(n=1)^N a_r)=sum_(n=1)^N a * r^(n-1) - sum_(n=2)^(N+1) a * r^(n-1)#

#color(white)((1-r) sum_(n=1)^N a_r)=a + color(red)(cancel(color(black)(sum_(n=2)^N a * r^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N a * r^(n-1)))) - a * r^N#

#color(white)((1-r) sum_(n=1)^N a_r)=a(1-r^N)#

Then (assuming

#sum_(n=1)^N a_r = (a(1-r^N))/(1-r)#

Note that if and only if

#sum_(n=1)^oo a_r = lim_(N->oo) sum_(n=1)^N a_r#

#color(white)(sum_(n=1)^oo a_r) = lim_(N->oo) (a(1-r^N))/(1-r)#

#color(white)(sum_(n=1)^oo a_r) = a/(1-r)#

So a geometric series has a convergent sum if and only if its common ratio is of absolute value less than

In our example with

#sum_(n=1)^oo (-8)(1/2)^(n-1) = (-8)/(1-1/2) = -16#