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mark these a_n b_n sequences watch

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    can someone please check if I did these right. I have no idea for 2a.

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    (Original post by cooldudeman)
    can someone please check if I did these right. I have no idea for 2a.

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    They look fine.

    For 2a, you could use n-1 >= n/2 if n >1, if you really want to get it into that final form.
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    (Original post by ghostwalker)
    They look fine.

    For 2a, you could use n-1 >= n/2 if n >1, if you really want to get it into that final form.
    thanks.
    what about this one:
    show that a_n tends to inf as n tends to inf and c is a positive real number, then c*a_n tends to inf as n tends to inf.

    it seems so obvious that this is true.
    all I can think of is if a_n --> infinity, then that given any A>0, a_n>A for all n>N so a_n >0 when n>N. we know c>0 and 2 positive numbers multiplied gives another positive so c*a_n tends to infinity.

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    (Original post by cooldudeman)

    show that a_n tends to inf as n tends to inf and c is a positive real number, then c*a_n tends to inf as n tends to inf.

    it seems so obvious that this is true.
    all I can think of is if a_n --> infinity, then that given any A>0, a_n>A for all n>N so a_n >0 when n>N. we know c>0 and 2 positive numbers multiplied gives another positive so c*a_n tends to infinity.

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    Your ending there seemed somewhat spurious.

    Let b_n = c*a_n

    Then so show b_n tends to infinity, given A, we need to find N such that b_n > A for n>N

    I.e. We need to find N such that c*a_n > A for n>N

    I.e. We need to find N such that a_n > A/c for n>N

    Now since a_n tends to infinty we can find such an N, for A/c. Hence, we have found our N for b_n.

    This general style of proof is worth noting.
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    (Original post by ghostwalker)
    Your ending there seemed somewhat spurious.

    Let b_n = c*a_n

    Then so show b_n tends to infinity, given A, we need to find N such that b_n > A for n>N

    I.e. We need to find N such that c*a_n > A for n>N

    I.e. We need to find N such that a_n > A/c for n>N

    Now since a_n tends to infinty we can find such an N, for A/c. Hence, we have found our N for b_n.

    This general style of proof is worth noting.
    um I need some time to soak this in lol. what did you mean by its worth nothing? are you saying that we don't have to follow a certain order when doing these questions?

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    (Original post by cooldudeman)
    um I need some time to soak this in lol. what did you mean by its worth nothing? are you saying that we don't have to follow a certain order when doing these questions?

    Posted from TSR Mobile
    It means I can't spell. Corrected it to "noting".

    There is an order to the method, which is hopefully clear.
 
 
 
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