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Thread: Infinite Series Proof.

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

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

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

    Then:
    I cosidered that sequence of the partial sums for the first seires $\displaystyle S_n=a_1+a_2+a_3+...+a_n$
    Since the series diverges, then { $\displaystyle S_n$ } diverges.
    i.e. $\displaystyle \lim_{n\to\infty} S_n = \infty$ , $\displaystyle -\infty$ or D.N.E
    The sequence of partial sums for the second series , denotes by $\displaystyle (S_n)^{'}$, is $\displaystyle (S_n)^{'} = cS_n$
    Clearly:
    $\displaystyle \lim_{n\to\infty} (S_n)^{'} = c \lim_{n\to\infty} S_n$
    if $\displaystyle \lim_{n\to\infty} S_n = \pm\infty$ then $\displaystyle \lim_{n\to\infty} (S_n)^{'} = \pm\infty$ depending on the sign of $\displaystyle c$
    Hence, The series $\displaystyle \sum_{n=1}^{\infty} ca_n$ diverges.
    But what if $\displaystyle \lim_{n\to\infty} S_n$ 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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    $\displaystyle \underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{ca_{n}}=c\underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{a_{n}}=c\sum\limits_{n=1 }^{\infty }{a_{n}},$ and the last series was given as divergent, so $\displaystyle \sum ca_n$ diverges as well.
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  3. #3
    Super Member General's Avatar
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle \underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{ca_{n}}=c\underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{a_{n}}=c\sum\limits_{n=1 }^{\infty }{a_{n}},$ and the last series was given as divergent, so $\displaystyle \sum ca_n$ 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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    $\displaystyle \lim_{n\to\infty}(-1)^n$ doesn't exist; what about $\displaystyle \lim_{n\to\infty}c(-1)^n,$ 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 $\displaystyle \sum_{n=1}^{\infty} a_n$ diverges and $\displaystyle c\neq0$ Show that :
    The series $\displaystyle \sum_{n=1}^{\infty} ca_n$ diverges also.

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

    Then:
    I cosidered that sequence of the partial sums for the first seires $\displaystyle S_n=a_1+a_2+a_3+...+a_n$
    Since the series diverges, then { $\displaystyle S_n$ } diverges.
    i.e. $\displaystyle \lim_{n\to\infty} S_n = \infty$ , $\displaystyle -\infty$ or D.N.E
    The sequence of partial sums for the second series , denotes by $\displaystyle (S_n)^{'}$, is $\displaystyle (S_n)^{'} = cS_n$
    Clearly:
    $\displaystyle \lim_{n\to\infty} (S_n)^{'} = c \lim_{n\to\infty} S_n$
    if $\displaystyle \lim_{n\to\infty} S_n = \pm\infty$ then $\displaystyle \lim_{n\to\infty} (S_n)^{'} = \pm\infty$ depending on the sign of $\displaystyle c$
    Hence, The series $\displaystyle \sum_{n=1}^{\infty} ca_n$ diverges.
    But what if $\displaystyle \lim_{n\to\infty} S_n$ 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 $\displaystyle S_m=\sum_{k=1}^{m}a_k$. Our assumption is that $\displaystyle \lim\text{ }S_m$ does not exist. Assume that $\displaystyle \lim\text{ }cS_m$ existed and equaled $\displaystyle L$. So, $\displaystyle \lim\text{ }cS_m=L$. Now, since $\displaystyle c\ne0$ and the left hand limit exists we see that $\displaystyle \lim\text{ }cS_m\cdot\lim\text{ }\frac{1}{c}=\lim\text{ }S_m=\frac{L}{c}$. Hence, we see that $\displaystyle \lim\text{ }S_m$ exists. Contradiction.
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