Question About Proof

  • Thread starter JG89
  • Start date
  • #1
726
1
In my calculus textbook there is a theorem:

"If a sequence a_1 + a_2 + ... converges and if b_1, b_2, ... is a bounded monotonic sequence of numbers, then (a_1)(b_1) + (a_2)(b_2) + ... converges"


Proof:

Let s_n denote the partial sums of [tex] \sum_{v=1}^n a_v [/tex], s the sum, and let [tex] \xi_n = s_n - s [/tex]. Then [tex] \sum_{v=n}^m a_v b_v = \sum_{v=n}^m (\xi_v - \xi_{v-1}) b_v = \sum_{v=n}^m \xi_v(b_v - b_{v+1}) - \xi_{n-1} b_n + \xi_m b_{m+1} [/tex].

For every sufficiently large v, [tex] |\xi_v| < \epsilon [/tex], and

[tex] \sum_{v=n}^m a_v b_v < \epsilon \sum_{v=n}^m |b_v - b_{v+1}| + \epsilon |b_n| + \epsilon |b_{m+1}| < \epsilon |b_n - b_{m+1}| + \epsilon |b_n| + \epsilon |b_{m+1}| [/tex].

This is in turn less than [tex] 4B \epsilon [/tex], where B is a bound for |b_v|, and the series [tex] \sum_{v=1}^{\infty} a_v b_v [/tex] converges

-------------------------------

I understand the proof and everything. I was wondering though, how did the writer of the proof know to rewrite the sum as this: [tex] \sum_{v=n}^m a_v b_v = \sum_{v=n}^m (\xi_v - \xi_{v-1}) b_v = \sum_{v=n}^m \xi_v(b_v - b_{v+1}) - \xi_{n-1} b_n + \xi_m b_{m+1} [/tex] ?

It just seems so random, something that I never would've thought about. If you could, could you please explain the thought processes he went through to realize he had to rewrite the sum in that form?

Thanks
 

Answers and Replies

  • #2
19
0
Hi JG89,

This is probably not the answer you were looking for. What you are describing is usually referred as Abel's test for convergence of the series in the form [tex]$\sum a_n b_n$[/tex]. The formula used in the proof is called Abel's partial summation formula. Every text that I've seen proves this theorem using this formula. It is a standard result that was discovered by Abel, probably by doing some other mathematics and uncovering this formula in the process. He was probably not trying to proof this theorem, since it didn't exist before him, or at least wasn't stated in these terms. This theorem probably followed the discovery of the formula, not the other way around.
 
  • #3
5
0
Hi JG89
With the definition of [tex] \xi_n [/tex] you can write [tex] \xi_n - \xi_{n-1} = s_n - s - (s_{n-1} - s) = s_n - s_{n-1} = \sum_{v=1}^n a_v - \sum_{v=1}^{n-1} a_v = a_n [/tex].
This gives the first step.
Now you can write the sum of the second step as follows:
[tex] \sum_{v=n}^m (\xi_v - \xi_{v-1})b_v = \sum_{v=n}^m \xi_vb_v - \sum_{v=n}^m \xi_{v-1}b_v = \sum_{v=n}^m \xi_v b_v - \sum_{k=n-1}^{m-1} \xi_k b_{k+1} = \sum_{v=n}^m \xi_v b_v- \sum_{k=n}^m \xi_kb_{k+1} - \xi_{n-1}b_n + \xi_mb_{m+1}. [/tex]
If you switch again to index v in the second sum you get the result. To do the second step in the above equation you set k=v-1, and then substitute every v in the sum.
 
  • #4
726
1
Hi JG89
With the definition of [tex] \xi_n [/tex] you can write [tex] \xi_n - \xi_{n-1} = s_n - s - (s_{n-1} - s) = s_n - s_{n-1} = \sum_{v=1}^n a_v - \sum_{v=1}^{n-1} a_v = a_n [/tex].
This gives the first step.
Now you can write the sum of the second step as follows:
[tex] \sum_{v=n}^m (\xi_v - \xi_{v-1})b_v = \sum_{v=n}^m \xi_vb_v - \sum_{v=n}^m \xi_{v-1}b_v = \sum_{v=n}^m \xi_v b_v - \sum_{k=n-1}^{m-1} \xi_k b_{k+1} = \sum_{v=n}^m \xi_v b_v- \sum_{k=n}^m \xi_kb_{k+1} - \xi_{n-1}b_n + \xi_mb_{m+1}. [/tex]
If you switch again to index v in the second sum you get the result. To do the second step in the above equation you set k=v-1, and then substitute every v in the sum.

Beautiful.

Thanks
 

Related Threads on Question About Proof

  • Last Post
Replies
3
Views
2K
Replies
1
Views
2K
Replies
2
Views
948
Replies
8
Views
2K
Replies
7
Views
609
  • Last Post
Replies
1
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
902
  • Last Post
Replies
4
Views
717
Top