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Topic: Difficulty with a proof in Baby Rudin
Replies: 4   Last Post: Apr 9, 2013 4:14 PM

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Paul

Posts: 385
Registered: 7/12/10
Re: Difficulty with a proof in Baby Rudin
Posted: Apr 9, 2013 4:54 AM
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On Monday, April 8, 2013 5:14:51 PM UTC+1, John Dawkins wrote:
> In article <d2e6896e-f170-4c9d-9fcf-56cbbb7bded6@googlegroups.com>,
>
> pepstein5@gmail.com wrote:
>
>
>

> > I have difficulty following the proof of theorem 8.3 in the 3rd edition of
>
> > Principles of Mathematical Analysis by Rudin. The line I'm struggling with
>
> > is:
>
> > lim n -> infinity [ sum(i = 1 to infinity) [sum j = 1 to n a_ij ] ] =
>
> > = lim n -> infinity [sum j = 1 to n [sum i = 1 to infinity a_ij ] ]
>
> >
>
> > This doesn't seem immediate, and the way I prove this is by theorem 3.55.
>
> > However, the whole statement of 8.3 seems immediate from 3.55 anyway, since
>
> > you just need to arrange the countable number of terms a_ij into a single
>
> > sequence and appeal to the absolute convergence.
>
>
>
> Rudin remarks on this point in the first sentence of his proof. But
>
> note that a double sum is not the same as a rearrangement of a single
>
> sum.
>
>
>

> > So Rudin's exposition of theorem 8.3 doesn't help me. I can only follow it
>
> > by using theorem 3.55 and I understand that his intention is to avoid theorem
>
> > 3.55.
>
> >
>
> > What am I missing?
>
>
>
> The series
>
>
>
> c_j := sum_{i = 1 to infinity} a_ij
>
>
>
> is absolutely convergent, by comparison to the series
>
> sum_{i = 1 to infinity} b_i. By the same token, the series
>
>
>
> d_n := sum_{i = 1 to infinity} [a_i1 + a_i2 + ... + a_in]
>
>
>
> is absolutely convergent, and
>
>
>
> d_n = sum_{i = 1 to infinity} sum_{j = 1 to n} a_ij
>
> = sum_{j = 1 to n} c_j.
>
>
>
> The limit lim_n d_n exists by the continuity of the function g
>
> introduced by Rudin in his proof; Because of the last display above,
>
> this in turn implies that the series sum_{j = 1 to infinity} c_j
>
> converges to lim_n d_n.
>
>

John,

Thank you very much for attending to this query. The good news is that I've been able to resolve this issue independently. However, I don't quite get your post.

I would imagine my difficulty is not unique so I'd like to repeat what the problem is and how I resolved it.

My problem was that I didn't see how to get:

lim n -> infinity [ sum(i = 1 to infinity) [sum j = 1 to n a_ij ] ] =
lim n -> infinity [sum j = 1 to n [sum i = 1 to infinity a_ij ] ]

My solution is: For all N

sum(i = 1 to N) [sum j = 1 to n a_ij ] =
sum j = 1 to n [sum i = 1 to N a_ij ]

Above is a rearrangement of finite sums.

Then take N -> infinity. Then take lim n -> infinity of both sides of the equation.

I think that leaving this argument out was a pedagogical mistake by Rudin, great though the book is. Yes, some readers would undoubtedly make the above steps with no special effort, but to assume that the typical reader would do so seems inconsistent with the elementary level of the text. At the very least, there should by a "Why?" accompanying that line to alert the reader to the gap.

Paul Epstein




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