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Topic: Analysis with series 1/2^2+1/3^2
Replies: 8   Last Post: Aug 9, 2012 12:17 PM

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 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: Analysis with series 1/2^2+1/3^2
Posted: Aug 9, 2012 12:11 PM

On Wed, 8 Aug 2012 11:37:16 -0500, "dilettante" <no@nonono.no> wrote:

>
>"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
>news:9p3528hcb1iaqmifjhlkcej9aij5mgagl6@4ax.com...

>> On Wed, 8 Aug 2012 07:23:02 -0700 (PDT), Mina <mina_world@hanmail.net>
>> wrote:
>>

>>>Hello teacher~
>>>
>>>{(1/2^2) + (1/3^2) + (1/4^2) + ...}
>>>+ {(1/2^3) + (1/3^3) + (1/4^3) + ...}
>>>+ {(1/2^4) + (1/3^4) + (1/4^4) + ...}
>>>+ ...
>>>
>>>----------------------------------------------
>>>I have a solution.(ambiguous)
>>>
>>>Namely,
>>>
>>>{(1/2^2) + (1/3^2) + (1/4^2) + ...}
>>>+ {(1/2^3) + (1/3^3) + (1/4^3) + ...}
>>>+ {(1/2^4) + (1/3^4) + (1/4^4) + ...}
>>>+ ...
>>>
>>>=
>>>
>>>(1/2^2 + 1/2^3 + 1/2^4 + ...)
>>>+ (1/3^2 + 1/3^3 + 1/3^4 + ...)
>>>+ (1/4^2 + 1/4^3 + 1/4^4 + ...)
>>>
>>>(this associative law ? Really possible?)

>>
>> It's not just the associative law. But this manipulation
>> is ok. The reason it's ok is because you're dealing with
>> a sum of _positive_ terms; if you have a sum of positive
>> terms any sort of regrouping or reordering is ok.

>
>Isn't it Fubini's theorem that justifies this, since we are not just
>rearranging the terms of a single series, but changing the order of a double
>summation?

Yes, it follows from Fubini's theorem. Or more precisely Tonelli's
theorem, which basically says "Fubini always works for positive
functions".

But the argument I had in mind is much more general, allowing
various manipulations that might not be covered by Tonelli.
Note I said "regrouping or reordering"...

Hmm. If S is any set, and a : S -> [0,infinty], let's define

int_S a = sup_F sum_{j in F} a(j),

where the sup runs over all finite sets F contained in S
(so sum_S a is just the integral of a with respect to
counting measure). And say N = {1,2,3...}.

Lemma 1. If a : N -> [0,infinity] then

sum_N a = sum_{j=1}^infinifty a(j).

Proof. Say s_n = sum_{j=1}^n a(j).
By definition of sum_N a it follows that

s_n <= sum_N a

for all n, hence

sum_{j=1}^infinity a(j) <= sum_N a.

Conversely, if F is a finite subset of N then
there exists A so that

s_n >= sum_F a (n > A),

and hence sum_{j=1}^infinity a(j) >= sum_N a. QED.

So order is irrelevant to sums of positive terms, since
it's clear that if phi is a permutation of S and a is a
positive function on S then

sum_S a = sum_S a o phi.

Similarly

Lemma 2. If (S_j)_{j in J} is any partition of S and
a is a positive function on S then

sum_S a = sum_j {sum_S_j a}.

Proof: Exercise; analogous to the previous. QED.

Those two lemmas suffice to show that

sum_j sum_k a(j,k) = sum_k sum_j a(j,k)

if a(j,k) >= 0.

My point in writing out the above is to point out that
the same argument applies to various other sorts
of "regrouping", for example

sum_j sum_k a(j,k)
= sum_n sum_j=1^{n-1} a(j, n-j).

I'm not going to try to define "any sort of regrouping
or reordering" - the same argument applies to
any sort of regrouping or reordering, where the
definition is "something where the same argument
applies"... As long as all the terms are positive
and each term appears exactly once there's
no problem.

>
>>
>>>
>>>=
>>>
>>>(1/2^2)/{1-(1/2)}
>>>+ (1/3^2)/{1-(1/3)}
>>>+ (1/4^2)/{1-(1/4)}
>>>+...
>>>
>>>= (1/2) + (1/3).(1/2) + (1/4).(1/3) + ...
>>>
>>>= (1/2) + {(1/2)-(1/3)} + {(1/3)-(1/4)} + ...
>>>
>>>= 1
>>>
>>>---------------------------------------------
>>>Hm... how do you think about it ?

>>

Date Subject Author
8/8/12 mina_world
8/8/12 dilettante
8/8/12 David C. Ullrich
8/8/12 dilettante
8/9/12 David C. Ullrich
8/8/12 Gottfried Helms
8/8/12 Phil H
8/9/12 Frederick Williams
8/9/12 David C. Ullrich