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Sum of Any Infinite Series


Date: 11/19/2000 at 02:00:28
From: Dan Keesing
Subject: If an Infinite Series Converges

Hi Dr. Math,

I just learned about various tests for convergence and the "powerful" 
Taylor series, but I was disappointed to learn that even if you can 
determine that a series converges, it's usually impossible to 
calculate an exact answer for the sum. Except for the rare cases like

     inf.
     SUM (1/n^2) = pi^2/6
     n=1

by Euler's proof, it would be extremely difficult to find an exact 
answer for an infinite sum. I am having trouble understanding the idea 
that if an infinite sum converges, it's not always possible to compute 
an actual answer - you can only get only a decimal or fractional 
approximation (using Taylor's theorem for the accuracy). Can you 
explain to me why it's not usually possible to compute this type of 
sum? Two examples of sums I'd like to find exactly are:

     inf.
     SUM 1/[(2n)!]
     n=1
and
     inf.
     SUM 10^n/[(n+1)*4^(2n+1)]
     n=1

By the Ratio Test, they both are absolutely convergent, so I'd like to 
see if it's possible to find an exact answer. Thanks!


Date: 11/21/2000 at 00:00:52
From: Doctor Fenton
Subject: Re: If an Infinite Series Converges

Dear Dan,

You've asked a very interesting question, and in some sense, the 
question is as much philosophical as mathematical. Part of the answer 
to your first question, why we can't find a "closed-form" solution to 
most series, in terms of "familiar" quantities, is connected with the 
question, "How many irrationals do you know?"

Irrational numbers fall into two camps, the algebraic numbers that are 
roots of polynomials with integer coefficients, such as sqrt(2), 
sqrt(3), etc., and transcendental numbers such as pi or e. If you know 
what a countable infinity is, then it will be meaningful to say that 
there are only countably many finite rational combinations of 
algebraic numbers, e and its powers e^2, e^3,... , pi and its powers, 
and rational roots of these, but there are uncountably many real 
numbers, so "most" real numbers cannot be expressed in terms of pi, e, 
or algebraic numbers to begin with, no matter how the real number 
arose. 

If you're not familiar with countability, there's a good discussion, 
"Set Theory and Orders of Infinity," here:

   http://mathforum.org/dr.math/problems/klimkow.4.8.97.html   

If you search our archives with a keyword like "uncountable" you'll 
find even more.

So, part of the problem is that you (and I) really don't know many 
irrational numbers by name. There are more real numbers than names, 
for that matter.

The philosophical part of the answer is the question why you feel 
more comfortable with an expression pi^2/6 than a decimal expansion. 
After all, you don't really know that many decimal places for pi, and 
why is writing the answer in terms of a quantity that has to be 
approximated any better than just giving an expression (the series) 
that will let you compute any desired number of decimal places. In 
fact, the known decimal places for pi and e come from similar series 
expansions. It's similar to the way most students feel more 
comfortable with degrees than with radians, even though the division 
of a circle into 360 parts is purely arbitrary, a 3000-year old legacy 
from the Babylonians, while radians are a "natural" definition of 
angles in terms of area or arclength. On the other hand, this question 
does bring up the question of why you should believe in the real 
numbers in the first place. Even though most people are mystified by 
i, the complex unit, real numbers are much more mystifying when you 
dig into them.

As for your series, those particular examples can actually be summed.
For example,

     e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ...
and
     1/e = 1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! + ...
so  
     (e + 1/e) = 2(1 + 1/2! + 1/4! + 1/6! + ...)

which means your first series is

     e + 1/e 
     -------  - 1
        2

The second series is tougher, but from the exponents on the 10 and 4, 
it resembles a geometric series. The (n+1) in the denominator can 
arise from integrating x^n, so I'd begin by trying to write this as

     inf.
     SUM (x^n)/(n+1)
     n=1

If I evaluate this series at x = 10/16, which is 10/4^2, I get your 
series, except for a factor of 4 in the denominator (we'll worry about 
that later.) Write

      1  inf. x^(n+1)    1  inf.  x
     --- SUM  ------- = --- SUM  INT t^n dt
      x  n=1   (n+1)     x  n=1   0  

                         1   x   inf.
                      = --- INT [SUM  t^n ] dt
                         x   0   n=1

                         1   x    t
                      = --- INT ----- dt
                         x   0   1-t

                         1
                      = --- [-x - ln(1-x)]
                         x 

Now divide this by 4 and evaluate this result at x = 10/16, and you'll 
have your second series. So, in fact, against the odds, the series you 
gave just happen to have "familiar" values.

If you have further questions, please write again.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Sequences, Series

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