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    if the limit of sequence a is infinity, and the limit of sequence b is c, how do i show the limit of sequence ab is infinity? also c>0
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    (Original post by asdfyolo)
    if the limit of sequence a is infinity, and the limit of sequence b is c, how do i show the limit of sequence ab is infinity? also c>0
    There are a few different ways to do it I should imagine, some brief thoughts on the subject - you'll want to be more rigorous than I have been.
    A brief sketch is that (b_n) converges so it is eventually close to c.
    Since this is the case, and you know that (a_n) is divergent, you can see that, for large n the sequence (a_nb_n) \sim c(a_n) and so must diverge. If you can express that more carefully, then you are done.

    Bigger hint in the spoiler - try to prove it yourself first. [spoiler]Eventually you have b_n>c-\epsilon.
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    An alternative approach: You may already have done some of these steps, in which case you can skip them.

    1) Prove that if c > 0, (a_n) a sequence diverging to +\infty, then (ca_n)\to 0.
    2) Prove that if (a_n) is a sequence diverging to infinity, and (b_n) is a sequence such that b_n > a_n for all sufficiently large n, then (b_n)\to 0.
    3) Since (b_n)\to c, there is in particular some N in the naturals such that for all n > N, b_n > c/2 > 0. Use this, and the above, to prove the result.
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    (Original post by joostan)
    There are a few different ways to do it I should imagine, some brief thoughts on the subject - you'll want to be more rigorous than I have been.
    A brief sketch is that (b_n) converges so it is eventually close to c.
    Since this is the case, and you know that (a_n) is divergent, you can see that, for large n the sequence (a_nb_n) \sim c(a_n) and so must diverge. If you can express that more carefully, then you are done.

    Bigger hint in the spoiler - try to prove it yourself first. [spoiler]Eventually you have b_n-\epsilon>c.
    Spoiler:
    Show
    This holds \forall \epsilon>0 for n sufficiently large.
    Spoiler:
    Show
    You'll now want to use the fact that a_n \rightarrow \infty - remember what this means.
    It is NOT true that eventually b_n - epsilon > c. Did you mean b_n > c - epsilon?
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    (Original post by DFranklin)
    It is NOT true that eventually b_n - epsilon > c. Did you mean b_n > c - epsilon?
    Yup, been having problems with spoilers, had to type it out twice - errors must've crept in.
    Thanks
 
 
 
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