Results 1 to 4 of 4

Math Help - Help me with proof

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    2

    Help me with proof

    Please dont laugh on my english if I make any mistake but im hungarian.
    Now let's see the problem:
    Proof if lim(an)=A and lim(bn)=B then lim(anbn)=AB
    Thank for your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,965
    Thanks
    1785
    Awards
    1
    Quote Originally Posted by zeleikristof View Post
    Please dont laugh on my english if I make any mistake but im hungarian.
    Now let's see the problem:
    Proof if lim(an)=A and lim(bn)=B then lim(anbn)=AB
    Thank for your help
    The sequence b_n is bounded: \left( {\exists M} \right)\left[ {\left| {b_n } \right| < M} \right].
    Then \left| {a_n b_n  - AB} \right| \leqslant \left| {a_n b_n  - Ab_n } \right| + \left| {Ab_n  - AB} \right| = \left| {b_n } \right|\left| {a_n  - A} \right| + \left| A \right|\left| {b_n  - B} \right|
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    From
    Johannesburg, South Africa
    Posts
    71
    Quote Originally Posted by zeleikristof View Post
    Please dont laugh on my english if I make any mistake but im hungarian.
    Now let's see the problem:
    Proof if lim(an)=A and lim(bn)=B then lim(anbn)=AB
    Thank for your help
    We have;
    \forall\ \epsilon>0\ \exists\ N\geq1\ \forall\ n\geq N \mid a_n-A \mid<\epsilon
    \forall\ \epsilon>0\ \exists\ M\geq1\ \forall\ l\geq M \mid b_l-B\mid<\epsilon
    We want to proof;
    \forall\ \epsilon>0\ \exists\ K\geq1\ \forall\ k\geq K \mid a_kb_k-AB\mid<\epsilon
    Start the proof as follow;
    \mid a_kb_k-AB\mid \leq \mid a_kb_k-a_kB+a_kB-AB\mid \leq \mid a_kb_k-a_kB\mid + \mid a_kB-AB\mid = \mid a_k(b_k-B)\mid + \mid B(a_k-A)\mid = \mid a_k \mid \mid b_k-B \mid + \mid B \mid \mid a_k-A \mid
    Use that a convergence sequence is bounded to proof the remaining part.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    2
    Thank you guys/girls!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 19th 2010, 11:50 AM
  2. Replies: 0
    Last Post: June 29th 2010, 09:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 11:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 02:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: April 14th 2008, 05:07 PM

Search Tags


/mathhelpforum @mathhelpforum