Subject: Infinite Geometric Series Who is asking: Student Level: Secondary Question: I ran into a problem when studying how to find the sum of an infinite geometric series. My math book attempts to explain the concept by giving formulas involving sigma and |r|, but it does not really explain how to go about finding the sum of an infinite geometric series. If you could either help me with this or point me in the direction of an informative website that could help me, I'd appreciate it. Thanks for your time. Sam Carter:)(: Hi Sam, I have two responses for you. One from Harley and one from Claude. The Centralizer. You have probably already seen the exprression for the sum of a finite geometric series but I want to develope it just to make sure we are using the same notation. Assume that a is positive and the sum of the finite geometric series is. Sn = a + ar + ar2 = ar3 +...+ arn-1 The usual technique is to multiply both sides by r to get rSn = ar + ar2 = ar3 +...+ arn-1 + arn Subtraction of the second equation from the first gives (1 - r)Sn = a(1 - rn) or To see what happens when n increases you need to consider four cases. I am not going to give a rigerous proof but rather try to illustrate the possibilities. Case 1. Suppose that -1 < r < 1, that is |r| < 1, for example r = 1/2 or r = -3/7. In this case as n increases rn gets close to zero. Hence as n increases, Sn approaches a/(1 - r) Case 2. Suppose that r < -1 or 1 < r, that is |r| > 1. If r > 1 then as n increases, rn increases in size without bound. If r < -1 then as n increases rn oscillates between positive and negative values but again increases in size without bound. In either situation as n incrreases Sn does not approach a specific value so we say that the sum of the infinite geometric series does not exist. Case 3. Suppose that r = 1 then the infinite series is a + a + a + a +... which again increases without bound, and hence the sum of the infinite geometric series does not exist. Case 4. Suppose that r = -1 then the infinite series is a - a + a - a +... which oscillaies between a and 0. Since the sum does not approach a specific value as n increases, we say that the sum of the infinite geometric series does not exist. In conclusion, if |r| < 1 then a + ar + ar2 = ar3 +...+ arn-1 +... = a/(1-r) Otherwise the infinite sum does not exist I hope this helps, Harley The "method" of finding the sum of an infinite geometric series is much more fun than the "formula". The trick is to find a way to have a repeating pattern, and then cancel it out. The classical example is the series 1/2 + 1/4 + 1/8 + 1/16 + ... where every term after the first is the previous term divided by 2. You want to find its sum X, so that you have 1/2 + 1/4 + 1/8 + 1/16 + ... = X To find a repetitive pattern, you divide everything by 2 (that is, you do to the whole series what you would normally do just to find the next term). (1/2 + 1/4 + 1/8 + 1/16 + ...)/2 = X/2 Distributing the division over the sum, you get 1/4 + 1/8 + 1/16 + 1/32 + ... = X/2 That is, if you start to add up at the second term of your series instead of the first, then you get an answer of X/2 instead of X. Therefore, the first term of the series was already X/2; so 1/2 = X/2, and X = 1. Thus you have 1/2 + 1/4 + 1/8 + 1/16 + .... = 1. In algebraic notation, you do as follows: Start with your series: 1/2 + 1/4 + 1/8 + 1/16 + ... = X Divide everything by 2: (1/2 + 1/4 + 1/8 + 1/16 + ...)/2 = X/2 Distribute the division: 1/4 + 1/8 + 1/16 + 1/32 + ... = X/2 Subtract the third line from the first: 1/2 + 1/4 + 1/8 + 1/16 + ... = X - ( 1/4 + 1/8 + 1/16 + 1/32 + ... = X/2 ) Or 1/2 + 0 + 0 + 0 + ... = X/2 And then solve for X: 1/2 = X/2, X = 1. This works also for other series; for instance, if you start at 5, and then add the terms obtained by dividing the previous term by 3, you have 5 + 5/3 + 5/9 + 5/27 + ... = X Divide everything by 3: (5 + 5/3 + 5/9 + 5/27 + ... )/3 = X/3 Distribute: 5/3 + 5/9 + 5/27 + 5/81 + ... = X/3 Subtract the third line from the first: 5 + 5/3 + 5/9 + 5/27 + ... = X - ( 5/3 + 5/9 + 5/27 + 5/81 + ... = X/3 ) Or 5 + 0 + 0 + 0 + ... = 2X/3 Then solve for X: 5 = 2X/3, 15 = 2X, 15/2 = X. Therefore you have: 5 + 5/3 + 5/9 + 5/27 + ... = 15/2. There is a logical and philosophical difficulty with this method: The three little dots represent an infinite continuation. So for instance, in the previous calculation, there are infinitely many subtractions: 5/3 - 5/3 = 0, 5/9 - 5/9 = 0, 5/27 - 5/27 = 0, and so on. Sure, these are all easy, but still you could never write them all down, so you are representing a calculation that you could never carry out completely. This can lead to strange things: If you start at 1 and form a series by MULTIPLYING the terms by 2, you get 1 + 2 + 4 + 8 + ... = X Multiply everything by 2: 2( 1 + 2 + 4 + 8 + ... = X ) Distribute: 2 + 4 + 8 + 16 + ... = 2X Subtract the third line from the first: 1 + 2 + 4 + 8 + ... = X - ( 2 + 4 + 8 + 16 + ... = 2X ) Or 1 + 0 + 0 + 0 + ... = -X Solve for X: -X = 1, X = -1. Therefore you have 1 + 2 + 4 + 8 + ... = -1. Surely this cannot be right! You start at 1, and keep on adding larger and larger positive numbers; how could you end up with a sum of -1? There has to be something wrong with this calculation! How about the first series we tried: 1/2 + 1/4 + 1/8 + 1/16 + ... = 1 This looks right: If I start with a pizza, eat half of it on the first day, then half of what is left on the second day, then half of what is left on the third day, and so on, then day by day I am eating 1/2 pizza + 1/4 pizza + 1/8 pizza + ... and it seems that over the course of eternity, I should get through my whole pizza. So, is the calculation correct or not? Difficulties like these stumped scientists for many centuries, and it is reasoning them through that opened the door to calculus and modern mathematics. Cheers, Claude Go to Math Central