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Absolute Convergence

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Absolutely Convergent Series

let _ \{ ~a_~i \}_{~i &in. &naturals.} _ be a sequence of real or complex numbers, _ then _ sum{ ~a_~i ,1,&infty.} _ is said to be _ #~{absolutely convergent} _ if _ sum{ | ~a_~i | ,1,&infty.} _ _ is convergent.

sum{ ~a_~i ,1,&infty.} _ is absolutely convergent _ => _ it is convergent.

Proof:

By _ Cauchy's general principle of convergence , _ given &epsilon. > 0 _ &exist. ~n_0 _ such that _ | ~t_~m - ~t_~n | < &epsilon. _ &forall. ~m , ~n > ~n_0 _ where _ ~t_~n = sum{ ~a_~i ,1,&infty.}.
I.e.

sum{ | ~a_~i | ,~i = ~n + 1,~n + ~k} _ < _ &epsilon. _ _ &forall. _ ~k >= 1, _ ~n > ~n_0

=> _ | sum{ ~a_~i ,~i = ~n + 1,~n + ~k} | _ < _ &epsilon. _ ( &triangle. inequality)

but

| sum{ ~a_~i ,~i = ~n + 1,~n + ~k} | _ = _ | ~s_{~n + ~k} - ~s_~n |

so

~s_~n _ rndb{ _ = _ sum{ ~a_~i ,~i = 1,~n}} _ is convergent, by Cauchy's general principle.

~{Alternatively}, _ define the sequences _ &set. ~a_~i^#{+} &xset. _ and _ &set. ~a_~i^#{-} &xset. _ by

~a_~i^#{+} _ = _ array{{~a_~i , _ _ _ ~a_~i >= 0}/{0 , _ _ _ ~a_~i < 0}} _ _ ; _ _ _ _ _ _ _ _ _ _ ~a_~i^#{-} _ = _ array{{0 , _ _ _ ~a_~i >= 0}/{-~a_~i , _ _ _ ~a_~i < 0}}

then _ ~a_~i = ~a_~i^#{+} + ~a_~i^#{-} , _ | ~a_~i | = ~a_~i^#{+} - ~a_~i^#{-} , _ and _ ~a_~i^#{+}, ~a_~i^#{-} >= 0 _ &forall. ~i.
~a_~i^#{+} =< | ~a_~i | _ &forall. ~i , _ ~a_~i^#{-} =< | ~a_~i | _ &forall. ~i _ ( remember that ~a_~i can be complex ), _ &therefore. _ &sum. ~a_~i^#{+} _ and _ &sum. ~a_~i^#{-} _ both convergent.
=> _ &sum. ~a_~i = &sum. ~a_~i^#{+} - &sum. ~a_~i^#{-} _ is also convergent.

Examples of Convergence

Exponential Series

~a_~n _ _ [ = _ _ ~a_~n(~x) ] _ = _ fract{~x^~n,~n#!} , _ _ _ _ ~n = 0, 1, 2, ...

sum{~a_~n,0,&infty.} _ = _ 1 + ~x + fract{~x^2,2#!} + fract{~x^3,3#!} + ...

det{fract{~a_{~n + 1} , ~a_~n }} = det{fract{ ~x , ~n }} _ -> _ 0 , _ as ~n -> &infty.

The exponential series is absolutely convergent &forall. real ~x.

Note that the number e as previously defined is given by:
_ _ _ _ _ e = &sum._0^{&infty.} ~a_~n(1) .

Binomial Series

Let _ ~m _ be an integer, _ ~m > 0, _ put

~a_~n _ = _ comb{~m,~n - 1}~x^{~n - 1}, _ _ _ _ ~n = 1, 2, ...

sum{~a_~n,1,&infty.} _ = _ 1 + ~m~x + fract{~m(~m - 1)~x^2,2} + fract{~m(~m - 1)(~m - 2)~x^3,3} + ...

det{fract{~a_{~n + 1} , ~a_~n }} = det{ fract{ ~m#! , (~m - ~n)#!~n#! } fract{ (~m - ~n + 1)#!(~n - 1)#!, ~m#! } fract{ ~x^~n , ~x^{~n -1 } } }

_ = _ det{ fract{ ~m - ~n + 1 , ~n } ~x } _ -> _ | ~x | , _ as ~n -> &infty.

The binomial series is absolutely convergent &forall. | ~x | < 1.

Logarithmic Series

~a_~n _ = _ (-1)^{~n+1} fract{~x^~n,~n } , _ _ _ _ ~n = 1, 2, ...

sum{~a_~n,1,&infty.} _ = _ ~x - fract{~x^2,2} + fract{~x^3,3} - fract{~x^4,4} + ...

det{fract{~a_{~n + 1} , ~a_~n }} = det{fract{ ~n~x , ~n + 1 }} _ -> _ | ~x | , _ as ~n -> &infty.

The logarithmic series is absolutely convergent &forall. | ~x | < 1.

Re-arrangement of Series Terms

Consider the alternate harmonic series.

~a_~n _ = _ (-1)^{~n+1} fract{1,~n } , _ _ _ _ ~n = 1, 2, ...

sum{~a_~n,1,&infty.} _ = _ 1 - fract{1,2} + fract{1,3} - fract{1,4} + ...

This was shown to be convergent to a number between &hlf. and 1. Now consider

sum{~a_~n,1,&infty.} _ = _ rndb{1 - fract{1,2}} - fract{1,4} + rndb{fract{1,3} - fract{1,6}} - fract{1,8} + ...

_ = _ fract{1,2} - fract{1,4} + fract{1,6} - fract{1,8} + ... _ = _ &hlf. sum{~a_~n,1,&infty.} _ #?

So how valid is it to re-arrange the terms of an infinite series?

#{Lemma}: _ If _ &sum. ~b_~n _ is a re-arrangement of (infinite) series _ &sum. ~a_~n , _ and _ &sum. ~a_~n _ is absolutely convergent, _ then _ &sum. ~a_~n _ and _ &sum. ~b_~n _ converge to the same limit.

Proof:
To be provided

Double Sequences

A #~{double sequence} is one that is indexed by two indices (from the natural numbers), _ i.e. one of the form _ &set. ~a_{~i , ~j} &xset._{~i &in. &naturals., ~j &in. &naturals.}
This can be represented by a two dimensional infinite array:

 

array{~a_{1,1} {,},~a_{1,2} {,},~a_{1,3} {,}, ... /~a_{2,1} {,},~a_{2,2} {,},~a_{2,3} {,}, ... /~a_{3,1} {,},~a_{3,2} {,},~a_{3,3} {,}, ... / ... , ... , ... , ... }

The sequence has a _ #~{sum by squares}:

~s _ = _ lim{~s_~n, ~n -> &infty.} _ = _ lim{rndb{ sum{,~i = 1,~n}sum{ ~a_{~i , ~j} ,~j = 1,~n}}, ~n -> &infty.}

providing the limit exists,

and a _ #~{sum by diagonals}:

~d _ _ = _ lim{~d_~n, ~n -> &infty.} _ = _ lim{rndb{sum{, _ ~j = 1,~n}sum{ ~a_{~i , ~j - ~i + 1} ,~i = 1, _ ~j}}, ~n -> &infty.}

_ = _ ~a_{1,1} _ _ + _ _ ~a_{1,2} + ~a_{2,1} _ + _ ~a_{1,3} + ~a_{2,2} + ~a_{3,1} _ _ + _ _ ...

_ = _ ( ~d_1 ) _ _ + _ _ ( _ _ ~d_2 _ _ ) _ _ + _ _ ( _ _ _ _ _ ~d_3 _ _ _ _ _ ) _ _ + ...

again, providing the limit exists.

#{Lemma}: _ If _ &set. ~a_{~i , ~j} &xset. is a sequence of non-negative terms, and either its sum by squares or sum by diagonals exist, then both sums exist and they are equal.

Proof:
Suppose _ ~s_~n &uparrow. ~s , _ ~s_~n = &sum._1^~n &sum._1^~n ~a_{~i ~j} . Then ~d_~n = &sum._1^~n &sum._1^~j ~a_{~i , ~j-1+~i} =< ~s_~n. _ So &set. ~d_~n &xset. is bounded above by ~s, _ i.e. _ ~d_~n &uparrow. ~d =< ~s.
Conversely ~s_{2~n} =< ~d_{2~n -1} , _ so _ ~s_~n &uparrow. ~s =< ~d.

#{Lemma}: _ If _ &set. | ~a_{~i , ~j} | &xset. has a sum by squares or sum by diagonals , then _ &set. ~a_{~i , ~j} &xset. _ has both sums and they are equal.

Proof:
| ~a_{~i , ~j} | = ~a_{~i , ~j}^#{+} + ~a_{~i , ~j}^#{-} _ (see definition above). _ &set. | ~a_{~i , ~j}^#{+} | &xset. _ and _ &set. | ~a_{~i , ~j}^#{-} | &xset. _ are non-negative and
~a_{~i , ~j}^#{+} =< | ~a_{~i , ~j} | , _ ~a_{~i , ~j}^#{+} =< | ~a_{~i , ~j} | , _ &forall. ~i, ~j.
So _ &set. | ~a_{~i , ~j}^#{+} | &xset. _ and _ &set. | ~a_{~i , ~j}^#{-} | &xset. _ have both sum by squares ( ~s^#{+} and ~s^#{-} ) and sum by diagonals ( ~d^#{+} and ~d^#{-} ), by above lemma, _ and _ ~a_{~i , ~j} _ = _ ~a_{~i , ~j}^#{+} - ~a_{~i , ~j}^#{-} _ -> _ ~s^#{+} - ~s^#{-} _ = _ ~d^#{+} - ~d^#{-}

Cauchy Product of Series

Consider the series _ &sum._1^{&infty.} ~a_~i _ and _ &sum._1^{&infty.} ~b_~j . _ Their _ #~{Cauchy product} _ is defined as:

sum{~c_~j, _ ~j = 0,&infty.} _ = _ sum{, _ ~j = 0,&infty.} sum{~a_~j ~a_{~j - ~i},~i = 0, _ ~j}

If _ &sum._1^{&infty.} ~a_~i _ and _ &sum._1^{&infty.} ~b_~j _ are absolutely convergent to the sums ~a and ~b respectively, then their Cauchy product is absolutely convergent to the sum ~a~b.

Proof:
Consider the double sequence _ &set. ~a_~i ~b_~j &xset. . _ Its sum by squares is

 

lim{rndb{ sum{,~i = 1,~n}sum{ ~a_~i ~b_~j , _ ~j = 1,~n}}, ~n -> &infty.} _ = _ lim{rndb{ sum{~a_~i,~i = 1,~n}rndb{ sum{~b_~j , _ ~j = 1,~n}}}, ~n -> &infty.}

_ = _ lim{rndb{ sum{~a_~i,~i = 1,~n}}rndb{ sum{~b_~j , _ ~j = 1,~n}}, ~n -> &infty.} _ = _ ~a~b

But the Cauchy product _ = _ sum by diagonals of _ &set. ~a_~i ~b_~j &xset. _ = _ sum by squares _ = _ ~a~b.

Power Series

A series of the form _ &sum._~n^{&infty.}_{= 0} _ ~a_~n ~x^~n , _ ~a_~n &in. &reals. , _ ~x &in. &reals. , _ is called a _ #~{power series}.

Radius of Convergence

By D'Alembert's ratio test _ &sum._~n^{&infty.}_{= 0} _ ~a_~n ~x^~n _ is absolutely convergent if

fract{ | ~a_{~n + 1} ~x^{~n + 1} | , | ~a_~n ~x^~n | } _ = _ | ~x | fract{ | ~a_{~n + 1} | , | ~a_~n | } _ _ -> _ _ ~k _ < _ 1

i.e. if _ _ _ | ~x | lim_{~n -> &infty.} ( | ~a_{~n + 1} / ~a_~n | ) _ < _ 1

or _ _ _ _ | ~x | _ < _ lim_{~n -> &infty.} ( | ~a_~n / ~a_{~n + 1} | ) _ =#: _ ~R

Then the series is said to have a #~{radius of convergence} ~R.

By D'Alembert, if _ | ~x | _ > _ ~R _ then the series is absolutely divergent, _ whereas the case | ~x | _ = _ ~R _ is inconclusive.