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

1 Answer
Jan 23, 2016

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)#