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JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Absolutely convergent series of rational numbers

Last edited by JaneFairfax (2007-04-11 03:27:07)

whatismath

Member

Registered: 2007-04-10

Posts: 19

Re: Absolutely convergent series of rational numbers

I think the answer is surely it can. Sorry I cannot think of a simpler construction
than the following. I will explain the idea after it.
We know the natural log base, e = 2.718281828459045... is irrational,
e = 1 + 1 + summation (n=2 to n=infinity) (1/n!)

and for n >=2, 1/n! = 1/(2.3.4...n) <= 1/(2.2..2) (n-1 times) = 1/(2^(n-1)).
So the series of non-negative rational numbers
summation (n=2 to n=infinity) (1/n!) = e -2
summation (n=2 to n=infinity) 1/(2^(n-1)) - (1/n!) = 1-(e-2) = 3-e

Consider the sequence a_n of "merging" the above and that of e:

a_0 = 1
a_1 = 1
a_n = 1/n!                        if n is even and n > 1
a_n = - (1/(2^(n-1)) - (1/n!))    if n is odd and n > 1

Then
  summation (n=0 to n=infinity) abs(a_n)
= 1+1+ (n=2 to n=infinity) ((1/n!) + 1/(2^(n-1)) - (1/n!))
= 1+1+1
=3 is rational

but
  summation (n=0 to n=infinity) a_n
= 1+1 + (n=2 to n=infinity) ((1/n!) - (1/(2^(n-1)) - (1/n!)))
= e - (3 -e)
= 2e -3 is irrational

The general idea is now that summation (n=0 to n=infinity) a_n converges,
we would like to find an example for which both
A = summation (n=0 to n=infinity) a_n with a_n >=0 and
B = summation (n=0 to n=infinity) a_n with a_n < 0 converge.

Then
     summation (n=0 to n=infinity) abs(a_n) = A - B
     summation (n=0 to n=infinity) a_n      = A + B

Pick any A, B such that A+B is rational but A-B is not, say
A = e = 2.718281828459045
B = 3-e = 0.28171...
as above.

The set of rational numbers being "dense" in the set of real numbers, meaning
there are sequences of rational numbers arbitrarily close to any given irrational
number and vice versa, we could construct series of rational numbers (even >=0)
converging to A and B respectively. Then "merge" the series to get {a_n}.
For example,

e     = 2 + .7 + .01 + .008 + .0002 + ... ( I use decimals for convenience)
3-e  =       .2 + .08 + .001 + .0007 + ...

3     = 2 + .7 + .2 + .01 + .08 +.008 + .001 + .0002 + .0007 + ...
2e-3= 2 + .7  - .2 + .01 - .08  +.008 - .001 + .0002 - .0007 + ... (answer)

Thanks.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Absolutely convergent series of rational numbers

Wow, thanks for the example! You just made my day.

This would that absolute convergence does imply convergence in

, although it does in

  and

. This is because

  and

  (considered as metric spaces) are complete, i.e. every Cauchy sequence converges.

, on the other hand, is not a complete metric space.