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Balance Point between Converging and Diverging Infinite Series


Date: 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/   
    
Associated Topics:
High School Calculus
High School Sequences, Series

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