Question

How do you find the sum of the infinite geometric series 1, 5, -25, 125,…?

Answer 1

This is not a geometric series.

If the first term is replaced with `-1` then the sum of the resulting geometric series does not converge.

Explanation:

If the sequence of numbers in the question is accurate then this is not a geometric series.

The ratios of successive pairs of terms are `5`, `-5`, `-5`.

It would be a geometric series if the first term was `-1`, with common ratio `-5`.

`-1, 5, -25, 125,...`

The general term of this sequence is `a_n = (-1) (-5)^(n-1)`

Then assuming that was the intention, the sum of this geometric series does not converge.

The sum of an infinite geometric sequence with general term `a_n = a r^(n-1)` converges when `abs(r) < 1` and not otherwise.

In general:

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

provided `abs(r) < 1`

We find:

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

`= sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)`

`= a + color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - ar^N`

`= a(1-r^N)`

Dividing both ends by `(1-r)` we get:

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

If `abs(r) < 1` then `lim_(N->oo) r^N = 0`

Hence:

`sum_(n=1)^oo a r^(n-1) = lim_(N->oo) sum_(n=1)^N a r^(n-1) = lim_(N->oo) (a(1-r^N))/(1-r) = a/(1-r)`