Sum of a Geometric ProgressionDate: 12/01/2002 at 07:55:22 From: Peter Subject: Infinite sum Dear Doctors, I found this strange equation on the internet, but didn't find any answer: 1 ----- = 1 + x + x^2 + x^3........ (1-x) It's not hard to verify that: (1-x)(1 + x + x^2 + x^3........) = 1 - x + x - x^2 + x^2..... = 1 Then this equation must be true for any number x <> 1 Let's try to plug the number 2 instead of x: 1 ----- = -1 = 1 + 2 + 4 + 8 + 16....... (1-2) It's really strange that an infinite sum of positive numbers is negative! Please could you help me in solving the problem or finding the error in the equation? Thanks for any help you provide. Sincerely, Peter Date: 12/01/2002 at 08:37:06 From: Doctor Jerry Subject: Re: Infinite sum Hi Peter, The series (1-x)^{-1} = 1+x+x^2+... converges only if |x| < 1. Otherwise, it does not. The series can be seen as the sum of a geometric progression. We know that a+a*r+a*r^2+... = a/(1-r), provided that |r| < 1. Here, a=1 and r=x. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 12/01/2002 at 08:43:40 From: Peter Subject: Infinite sum Dear Dr. Jerry, Thanks for responding to my question, but unfortunately, I didn't understand your answer. 1) why does (1-x)^(-1)=1+x+x^2.... converge only if |x|<1 ? 2) why does a+a*r+a*r^2+... = a/(1-r) provided that |r|<1 ? Thanks for any help, Peter Date: 12/01/2002 at 15:12:18 From: Doctor Jerry Subject: Re: Infinite sum Hi Peter, An infinite series a1+a2+a3+... of numbers is said to converge provided that there is a number S such that for every positive number E there is a positive integer N for which |(a1+a2+...+an) - S | < E whenever n >= N. The number S is called the sum of the series. This definition is a standard definition of convergence. You can find it in most calculus books. For the series 1+x+x^2+... it is clear that it does not converge to a number S when x = 1. The sums 1+x+...+x^{n-1} = n and hence grow beyond all bounds as n increases. We can form the partial sum 1+x+...+x^{n-1} of the first n terms as follows: denoting this partial sum by Sn, Sn = 1+x+...+x^{n-1} x*Sn = x+x^2+...+x^n Sn - x*Sn = 1 - x^n Hence, Sn = (1-x^n)/(1-x). If |x|<1, we see that the limit of the numerator is 1. Hence, Sn approaches 1/(1-x) as n increases without bound. In the definition of convergence set out above, we take S=1/(1-x); in this case, |S-(1+x+x^2+...+x^{n-1}| = |1/(1-x) - (1-x^n)/(1-x)| = |x^n/(1-x)| With x fixed and |x| < 1, if E is any positive number, all we have to do is to take N sufficiently large that |x^N| < E(1-x). The rest follows easily. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/