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Math Help - counter examples in real analysis

  1. #1
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    counter examples in real analysis

    For each of the following, prove or give a counter example:
    a) if (Sn) converges to S, then (|Sn|) converges to |S|
    b) if (|Sn|) is convergent, then (Sn) is convergent
    c) lim Sn = o iff lim |Sn| = 0
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  2. #2
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    Quote Originally Posted by learn18 View Post
    For each of the following, prove or give a counter example:
    a) if (Sn) converges to S, then (|Sn|) converges to |S|
    b) if (|Sn|) is convergent, then (Sn) is convergent
    c) lim Sn = o iff lim |Sn| = 0
    b) counter example Sn=(-1)^n, then |Sn| converges to 1, and Sn does not converge

    RonL
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  3. #3
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    Quote Originally Posted by learn18 View Post
    For each of the following, prove or give a counter example:
    a) if (Sn) converges to S, then (|Sn|) converges to |S|
    b) if (|Sn|) is convergent, then (Sn) is convergent
    c) lim Sn = o iff lim |Sn| = 0
    c) Suppose lim(n->inf) Sn = 0, this means that for all epsilon in R there exists
    a natural number N such that:

    |Sn - 0| < epsilon, for all n>N.

    but this last is the same thing as:

    |Sn| < epsilon, for all n>N

    which is again the same as:

    ||Sn|-0| < epsilon, for all n>N

    so lim(n->inf) Sn = 0 implies that lim(n->inf) |Sn| = 0.


    For the other half of the proof, suppose lim(n->inf) |Sn| = 0, this means that
    for all epsilon in R there exists a natural number N such that:

    ||Sn| - 0| < epsilon, for all n>N.

    but this last is the same thing as:

    |Sn| < epsilon, for all n>N

    which is again the same as:

    |Sn-0| < epsilon, for all n>N

    so lim(n->inf) |Sn| = 0 implies that lim(n->inf) Sn = 0.

    Together these proove: lim Sn = 0 iff lim |Sn| = 0

    RonL
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  4. #4
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    Quote Originally Posted by learn18 View Post
    For each of the following, prove or give a counter example:
    a) if (Sn) converges to S, then (|Sn|) converges to |S|
    if (s_n) converges to s then,
    |s_n - s| < e for n>N

    Note that,
    ||x|-|y||<=|x-y|

    Thus,
    ||s_n|-|s||<=|s_n-s|<e for n>N

    Thus,
    lim |s_n| = |s|
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