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  1. #1
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    help with sequences

    Prove:
    lim (KSn) = KS and lim (K + Sn) = K + S, for any k element of R

    Sn means S sub n
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by slowcurv99 View Post
    Prove:
    lim (KSn) = KS and lim (K + Sn) = K + S, for any k element of R

    Sn means S sub n
    So first we need to define what it means to be a limit. There ar etwo major definitions, which are essentially the same. I have used both, the first from Calc 1 to calc 3 and the latter from the proofs class i'm doing now. I will prove the first question using both methods, so you can see how they are similar and different, and the last by the latter definition only. Here are the definitions:

    Definition 1:

    Let be a function defined on an open interval containing (except possibly at ) and let be a real number. means that
    for each real there exists a real such that for all where we have .
    Definition 2:

    We say the limit of a sequence (Sn) is a real number, s, provided that for each e > 0 there exists a natural number N such that n>N implies |Sn - S| < e

    Note: e means epsilon. d means delta


    lim (KSn) = KS
    Proof (By definition 1): The result is obvious for k = 0. Assume k is not 0. Let limSn = S. So for each e>0 there exists a d>0 such that 0<|x - c|<d implies |Sn - S|<e/|k|.

    Then 0<|x - c|<d implies |kSn - kS| <or= |k||Sn - S| < |k|.e/|k| = e

    So we have for each e>0, there is a d>0 such that 0<|x - c|<d implies |kSn - kS|< e. Thus the limkSn = kS

    QED

    Alternate Proof (By definition 2): The result is obvious for k = 0, therefore, let us assume k is not 0.

    Let e>0. Since limSn = S, there exists an N such that n>N implies |Sn - S|< e/|k|

    So n>N implies |kSn - kS| <or= |k||Sn - S| < |k|.e/|k| = e

    Since n>N implies |kSn - kS| < e, limkSn = kS.

    QED





    lim (K + Sn) = K + S
    Proof (By Definition 2): The result is obvious for k = 0, so let's assume k is not 0. Let the limSn = S. This means for any e > 0, there exists an N such that n > N implies |Sn - S| < e

    Then n>N implies |(K + Sn) - (K + S)| = |Sn - S + K - K| = |Sn - S| < e

    Since n>N implies |(K + Sn) - (K + S)| < e, lim(K + Sn) = K + S

    QED

    The proof by definition 1 is very similar as you can see in the first question I did. You should be able to reproduce this result using that definition if that's the one you learned in class.
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    So first we need to define what it means to be a limit. There ar etwo major definitions, which are essentially the same. I have used both, the first from Calc 1 to calc 3 and the latter from the proofs class i'm doing now. I will prove the first question using both methods, so you can see how they are similar and different, and the last by the latter definition only. Here are the definitions:
    .

    These are not limits of functions, it is limits of sequences.

    lim (KSn) = KS
    Let me doth this one, and leave ye other to thee.

    Let {s_n} be a convergent sequence.
    Define lim s_n =s.

    There are two cases to consider.

    Case 1, k=0
    There is nothing to prove because k*s_n=0 which is a konstant sequence.
    Q.E.D.

    Case 2, k !=0
    For any e>0 we have, e/|k|>0.
    Thus, there exists an N such that,
    |s_n - s|<e/|k|
    Multiply by |k| to get,
    |s_n-s|*|k|=|ks_n-ks|<e
    Thus, {ks_n} is convergent and lim k*s_n=ks.

    Q.E.D.
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