Turn on thread page Beta
    • Thread Starter
    Offline

    15
    ReputationRep:
    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 (a_{n})_{n=1}^{\infty} be a sequence of real numbers. Suppose that each of the subsequences:
    (1)(a_{3n})_{n=1}^{\infty} \\

(2)(a_{3n+1})_{n=0}^{\infty} \\

(3) (a_{3n+2})_{n=0}^{\infty}
    converges to the limit a. Prove that the original sequence (a_{n}) converges to a.

    My attempt:
    (1) implies |a_{3n}-a|<\epsilon for all n>N_{0}
    (2) implies |a_{3n+1}-a|<\epsilon for all n>N_{1}
    (3) implies |a_{3n+2}-a|<\epsilon for all n>N_{2}

    Using the first result there exists some n>N_{0} such that the series converges to a when n is a multiple of 3.
    Using the second result there exists some n>N_{1} such that the series converges to a when n is a multiple of 3 plus 1.
    Using the third result there exists some n>N_{2} 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 n>max(N_{0},N_{1},N_{2}) we have |a_{n}-a|<\epsilon and so a_{n} converges to a.
    Offline

    12
    ReputationRep:
    I guess it is ok as you proved by definition
    Offline

    2
    ReputationRep:
    (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 (a_{n})_{n=1}^{\infty} be a sequence of real numbers. Suppose that each of the subsequences:
    (1)(a_{3n})_{n=1}^{\infty} \\

(2)(a_{3n+1})_{n=0}^{\infty} \\

(3) (a_{3n+2})_{n=0}^{\infty}
    converges to the limit a. Prove that the original sequence (a_{n}) converges to a.

    My attempt:
    (1) implies |a_{3n}-a|<\epsilon for all n>N_{0}
    (2) implies |a_{3n+1}-a|<\epsilon for all n>N_{1}
    (3) implies |a_{3n+2}-a|<\epsilon for all n>N_{2}

    Using the first result there exists some n>N_{0} such that the series converges to a when n is a multiple of 3.
    Using the second result there exists some n>N_{1} such that the series converges to a when n is a multiple of 3 plus 1.
    Using the third result there exists some n>N_{2} 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 n>max(N_{0},N_{1},N_{2}) we have |a_{n}-a|<\epsilon and so a_{n} 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.
    • Thread Starter
    Offline

    15
    ReputationRep:
    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.
    Offline

    2
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    15
    ReputationRep:
    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.
 
 
 
Turn on thread page Beta
Updated: July 4, 2005

1,410

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.