Question

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

Answer 1

`-8-4-2-... = -16`

Explanation:

Note that the given series is a geometric series with initial term `-8` and common ratio:

`(-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 `a` is the initial term and `r` the common ratio.

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 `r != 1`) dividing both ends by `(1-r)` we find:

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

Note that if and only if `abs(r) < 1` then `lim_(N->oo) r^N = 0` and we find:

`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 `1`.

In our example with `a=-8` and `r=1/2` we find:

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