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Math Help - Infinite Series Proof.

  1. #1
    Super Member General's Avatar
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    Infinite Series Proof.

    If \sum_{n=1}^{\infty} a_n diverges and c\neq0 Show that :
    The series \sum_{n=1}^{\infty} ca_n diverges also.

    First I thought about nth terms test, but it failed.
    since \lim_{n\to\infty} a_n can be zero.

    Then:
    I cosidered that sequence of the partial sums for the first seires S_n=a_1+a_2+a_3+...+a_n
    Since the series diverges, then { S_n } diverges.
    i.e. \lim_{n\to\infty} S_n = \infty , -\infty or D.N.E
    The sequence of partial sums for the second series , denotes by (S_n)^{'}, is (S_n)^{'} = cS_n
    Clearly:
    \lim_{n\to\infty} (S_n)^{'} = c \lim_{n\to\infty} S_n
    if \lim_{n\to\infty} S_n = \pm\infty then \lim_{n\to\infty} (S_n)^{'} = \pm\infty depending on the sign of c
    Hence, The series \sum_{n=1}^{\infty} ca_n diverges.
    But what if \lim_{n\to\infty} S_n D.N.E ?

    i.e. If you multiply a D.N.E Limit with by a constant, It will be D.N.E too ?
    My proof is right?
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  2. #2
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    \underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{ca_{n}}=c\underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{a_{n}}=c\sum\limits_{n=1  }^{\infty }{a_{n}}, and the last series was given as divergent, so \sum ca_n diverges as well.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    \underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{ca_{n}}=c\underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{a_{n}}=c\sum\limits_{n=1  }^{\infty }{a_{n}}, and the last series was given as divergent, so \sum ca_n diverges as well.
    Thanks.
    But what about my proof ?
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  4. #4
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    I see no problem with it.

    As for if the limit doesn't exist, it's nothing, that's all.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    I see no problem with it.

    As for if the limit doesn't exist, it's nothing, that's all.
    Thanks.
    So the limit which obtained by multiplying a D.N.E limit with a constant is D.N.E also ?
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  6. #6
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    \lim_{n\to\infty}(-1)^n doesn't exist; what about \lim_{n\to\infty}c(-1)^n, the conclusion is the same, it doesn't exist.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by General View Post
    If \sum_{n=1}^{\infty} a_n diverges and c\neq0 Show that :
    The series \sum_{n=1}^{\infty} ca_n diverges also.

    First I thought about nth terms test, but it failed.
    since \lim_{n\to\infty} a_n can be zero.

    Then:
    I cosidered that sequence of the partial sums for the first seires S_n=a_1+a_2+a_3+...+a_n
    Since the series diverges, then { S_n } diverges.
    i.e. \lim_{n\to\infty} S_n = \infty , -\infty or D.N.E
    The sequence of partial sums for the second series , denotes by (S_n)^{'}, is (S_n)^{'} = cS_n
    Clearly:
    \lim_{n\to\infty} (S_n)^{'} = c \lim_{n\to\infty} S_n
    if \lim_{n\to\infty} S_n = \pm\infty then \lim_{n\to\infty} (S_n)^{'} = \pm\infty depending on the sign of c
    Hence, The series \sum_{n=1}^{\infty} ca_n diverges.
    But what if \lim_{n\to\infty} S_n D.N.E ?

    i.e. If you multiply a D.N.E Limit with by a constant, It will be D.N.E too ?
    My proof is right?
    Let S_m=\sum_{k=1}^{m}a_k. Our assumption is that \lim\text{ }S_m does not exist. Assume that \lim\text{ }cS_m existed and equaled L. So, \lim\text{ }cS_m=L. Now, since c\ne0 and the left hand limit exists we see that \lim\text{ }cS_m\cdot\lim\text{ }\frac{1}{c}=\lim\text{ }S_m=\frac{L}{c}. Hence, we see that \lim\text{ }S_m exists. Contradiction.
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