Balance Point between Converging and Diverging Infinite SeriesDate: 03/20/2002 at 00:58:39 From: Jim Subject: Balance point between converging and diverging infinite series I have developed an interest in infinite series in an attempt to understand infinity in general. At first I thought that if the n(th) element in a series tended to zero as n tended to infinity, that the series would have to converge. However, this does not take into consideration that there is an infinite number of terms. The divergence of the harmonic series is a counter example to my original assumption. It seems to matter how quickly the n(th) term of an infinite series tends toward zero. The series 1/2 + 1/4 + 1/8 + ... is an example of a series where the n(th) term gets very small very quickly. This series converges. My question is: "Is there a way to find the balance point between convergence and divergence for any type of series? That point where the n(th)term gets smaller JUST fast enough for the series to converge." (Or am I making a wrong assumption about the nature of convergence?) Thanks. Date: 03/20/2002 at 10:35:11 From: Doctor Mitteldorf Subject: Re: Balance point Dear Jim, With your experimentation, you've created an understanding for yourself that (I'm guessing) no amount of textbook learning could provide. You have an excellent feel for what's going on. How fast the terms get small is exactly the crucial question here. To define a "balance point" you'll have to specify more precisely the relation among the numbers in your series. Here are some specific examples: The harmonic series is defined by a common ratio a between successive terms. If a<1, the series converges; for a>=1, the series diverges. It is curious that if a=1, the series turns into an infinite sum of terms that are all the same, so it obviously diverges; but if a=0.999, the series converges (to 1000 times the first term). The sum of the reciprocals of the integers 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... diverges very slowly. The sum after n terms is proportional to log(n). Clearly the sum of the reciprocals of the odd or even integers is still going to diverge. Curiously, the sum of all prime numbers, even though they get thinner and thinner, still diverges - very, very slowly. The sum of the reciprocals of all perfect squares, however, converges. (Can you find the value of this sum?) Did you have a specific functional form in mind when you asked about a "balance point" beyond which a series diverges? - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 03/20/2002 at 16:33:02 From: Jim Subject: Balance point Dr. Mitteldorf, Thank you for your very prompt reply to my question. I did some experimentation on the series 1 + 1/2^2 + 1/3^3 + 1/4^2 +... and found that the partial sum seems to converge somewhat short of sqrt(e). Some further searching in the Dr. Math archives led to the answer that the sum of this series is actually pi^2/6, although I do not understand the proof. Back to my original question.... The series 1 + 1/2 + 1/3 + 1/4 + ... diverges. The series 1 + 1/2^2 + 1/3^2 + 1/4^2 + ... converges. Let's consider a series of the form 1 + 1/2^r + 1/3^r + 1/4^r + ..., where r is between 1 and 2 (the two extremes being the previous examples. Question: Is there a real r between 1 and 2 for which the series 1 + 1/2^r + 1/3^r + 1/4^r + ... is at the balance point between convergence and divergence? Thanks for your help. Jim Date: 03/20/2002 at 17:42:21 From: Doctor Mitteldorf Subject: Re: Balance point Dear Jim- You've done some good sleuthing, I see. Are you familiar with integral calculus? The sum of numbers (1/n)^r is closely related to the integral of (1/x)^r. The integral is much easier to calculate, and might give you a hint about where the "balance point" is for r. - Doctor Mitteldorf, 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/