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# Basic analysis proof. watch

1. I have the following question and i'm not sure whether my 'proof' is correct as it seems much too wordy. What do I need to alter?

Let be a sequence of real numbers. Suppose that each of the subsequences:

converges to the limit a. Prove that the original sequence converges to a.

My attempt:
(1) implies for all
(2) implies for all
(3) implies for all

Using the first result there exists some such that the series converges to a when n is a multiple of 3.
Using the second result there exists some such that the series converges to a when n is a multiple of 3 plus 1.
Using the third result there exists some such that the series converges to a when n is a multiple of 3 plus 2.

But all positive integers are of the form 3n, 3n+1 or 3n+2 where n is another integer. Hence for all we have and so converges to a.
2. I guess it is ok as you proved by definition
3. (Original post by Gaz031)
I have the following question and i'm not sure whether my 'proof' is correct as it seems much too wordy. What do I need to alter?

Let be a sequence of real numbers. Suppose that each of the subsequences:

converges to the limit a. Prove that the original sequence converges to a.

My attempt:
(1) implies for all
(2) implies for all
(3) implies for all

Using the first result there exists some such that the series converges to a when n is a multiple of 3.
Using the second result there exists some such that the series converges to a when n is a multiple of 3 plus 1.
Using the third result there exists some such that the series converges to a when n is a multiple of 3 plus 2.

But all positive integers are of the form 3n, 3n+1 or 3n+2 where n is another integer. Hence for all we have and so converges to a.
Your blue paragraph is redundant, and the red paragraph is slightly wrong. You can replace the latter with the following.

Suppose that n > 3max{N0, N1, N2} + 2. Write n = 3k + i, where i is 0, 1 or 2. Then 3k + i > 3max{N0, N1, N2} + 2, so 3k > 3max{N0, N1, N2}, so k > max{N0, N1, N2}. Now we prove that |a_n - a| < epsilon by considering the three possible values of i.

Case 1: i = 0
|a_n - a|
= |a_(3k) - a|
< epsilon (because k > N0)

Case 2: i = 1
|a_n - a|
= |a_(3k + 1) - a|
< epsilon (because k > N1)

Case 3: i = 2
|a_n - a|
= |a_(3k + 2) - a|
< epsilon (because k > N2)

So in any case we have |a_n - a| < epsilon.

So a_n -> a as n -> infinity.
4. Thanks for the help.
I can understand how my blue paragraph is redundant.
However, why is it necessary to consider the inequality for n case by case depending on whether it is of the form 3k, 3k+1 or 3k+2? Surely n>max(N_0,N_1,N_2) covers all cases? Is the change just in order to explain things more or is there something in the red paragraph that I should keep in mind not to do anymore?
Thanks.
5. (Original post by Gaz031)
Thanks for the help.
I can understand how my blue paragraph is redundant.
However, why is it necessary to consider the inequality for n case by case depending on whether it is of the form 3k, 3k+1 or 3k+2? Surely n>max(N_0,N_1,N_2) covers all cases? Is the change just in order to explain things more or is there something in the red paragraph that I should keep in mind not to do anymore?
Thanks.
Suppose for example that N0 = 5, N1 = 6, N2 = 7.

Then we know:

|a_n - a| < epsilon for n = 18, 21, 24, ...
|a_n - a| < epsilon for n = 22, 25, 28, ...
|a_n - a| < epsilon for n = 26, 29, 32, ...

--

max{N0, N1, N2} = 7,
but n > 7 does not imply |a_n - a| < epsilon.

(So your red paragraph is wrong.)

--

3max{N0, N1, N2} + 2 = 23,
n > 23 implies |a_n - a| < epsilon.

--

You might prefer to say "n > max{3N0, 3N1 + 1, 3N2 + 2}" rather than "n > 3max{N0, N1, N2} + 2". Both give correct proofs.
6. Aah i'm neglecting the difference between a_(n) and a_(3n+2) (ie for n>n2 we need n>3n2+2 etc.) I need to remember to compare the term in the original sequence (ie the subscript) rather than the value of n.

You might prefer to say "n > max{3N0, 3N1 + 1, 3N2 + 2}" rather than "n > 3max{N0, N1, N2} + 2". Both give correct proofs.
Yes I probably prefer that.

Thanks, I understand now.

Heh, the next question is for the general set of subsequences a_(pn),a_(pn+1)...+...+a_(pn+p-1) causing the sequence to converge. Thankfully I won't make the same error now.

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