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# Worked example: convergent geometric series

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.3 (EK)
,
LIM‑7.A.4 (EK)

## Video transcript

let's get some practice taking sums of infinite geometric series so we have one over here and just to make sure that we're dealing with the geometric series let's make sure we have a common ratio so let's see to go from the first term to the second term we multiply by 1/3 and then go to the next term we are going to multiply by 1/3 again and we're going to keep doing that so we can rewrite this series as 8 plus 8 times 1/3 8 times 1/3 plus plus 8 times 1/3 squared 8 times 1/3 squared each successive term we multiply by 1/3 again and so when you look at it this way you're like okay we could write this in Sigma notation this is going to be equal to so this the first thing we wrote is equal to this which is equal to this is equal to the sum the sum and we could start at 0 or 1 depending on how we'd like to do it we could say from K is equal to 0 and this is an infinite series right here we're just going to keep on going forever so to infinity of well what's our first term our first term is 8 so it's going to be 8 times our common ratio times our common ratio 1 3 to the K to the K power and let me just verify that this indeed works and I always do this just as a reality check and I encourage you to do the same so when K equals a 0 that should be the first term right over here you get 8 times 1/3 to the 0 power which is indeed 8 when K is equal to 1 that's going to be our second term here that's going to be 8 times 1/3 to the first power that's what we have here and so when K is equal to a 2 that is this term right over here so the these are all describing the same thing so now that we've seen that we can write a geometric series in multiple ways let's find the sum well we've seen before and we proved it in other videos if you have a sum from k equals 0 to infinity and you have you or first-term a times R to the K power R to the K power assuming this converges so assuming that the absolute value of your common ratio is less than one this is what you need needs to be true for convergence this is going to be equal to this is going to be equal to our first term which is a over over 1 minus our common ratio 1 minus our common ratio and if this looks unfamiliar to you I encourage you to watch the video where we where we find the formula derived the formula for the sum of an infinite geometric series but just applying that over here we are going to get we are going to get this is going to be equal to our first term which is 8 so that is 8 over 1 minus 1 minus our common ratio over 1/3 and we know this is going to converge because our common ratio the the magnitude the absolute value of 1/3 is indeed less than 1 and so this is all going to converge to this is going to converge to 8 over 1 minus 1/3 is 2/3 2/3 which is the same thing as 8 times 3 halves which is let's see this could become divide 8 by 2 that becomes 4 and so this is this will become 12