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Thread: Determining whether a sequence of partial sums is convergent or divergent

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    Determining whether a sequence of partial sums is convergent or divergent

    Given the following sequence of partial sums:
    $\displaystyle S_N=\frac{1}{N}cos(N\pi)$

    My attempt:
    If I can find the series for this sequence of partial sums, then I can test the series to see if it's convergent or divergent.

    $\displaystyle a_N=S_N-S_{N-1}=\sum^N_{n=1}a_n-\sum^{N-1}_{n=1}a_n$

    Thus:
    $\displaystyle a_N=\frac{1}{N}cos(N\pi)-\frac{1}{N-1}cos((N-1)\pi)$

    Since $\displaystyle cos(N\pi)=(-1)^N$:

    $\displaystyle a_N=\frac{1}{N}(-1)^N-\frac{1}{N-1}(-1)^{N-1}$

    But this is as far as I got.
    I would like to express this with a factor $\displaystyle (-1)^N$ so that I can use the alternating series test, but I'm not sure how exactly.

    $\displaystyle a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$
    Would this be correct?
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    Quote Originally Posted by MathIsOhSoHard View Post
    Given the following sequence of partial sums:
    $\displaystyle S_N=\frac{1}{N}cos(N\pi)$
    If the real question is contained in the title of this thread, then you have wasted a lot effort.
    Because the partial sums are given as $\displaystyle S_N=\frac{1}{N}cos(N\pi)$.
    You know $\displaystyle (S_N)\to 0$ so the series converges to zero.

    If the question asks something else, then the title is misleading.
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    This is the question in its entirety:

    For a series $\displaystyle \sum^{/infty}_{n=1}a_n$ the sequence of partial sums is $\displaystyle S_N=\frac{1}{N}cos(N\pi)$. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
    Hint: Find the series n'th term $\displaystyle a_n$ and express it as an alternating series.


    (It's supposed to be an infinity sign above the sum, but for some reason LATEX doesn't want to display it)
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    Quote Originally Posted by MathIsOhSoHard View Post
    This is the question in its entirety:
    For a series $\displaystyle \sum^{/infty}_{n=1}a_n$ the sequence of partial sums is $\displaystyle S_N=\frac{1}{N}cos(N\pi)$. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
    Hint: Find the series n'th term $\displaystyle a_n$ and express it as an alternating series.
    Well the only difficult part is "Determine whether the series is absolutely convergent,"
    For that reason you may want to find $\displaystyle a_n$.
    $\displaystyle a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$.

    [tex]\sum\limits_{n = 0}^\infty {ar^n } [/tex] gives $\displaystyle \sum\limits_{n = 0}^\infty {ar^n } $
    Last edited by Plato; Nov 2nd 2012 at 12:34 PM.
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    Quote Originally Posted by Plato View Post
    Well the only difficult part is "Determine whether the series is absolutely convergent,"
    For that reason you may want to find $\displaystyle a_n$.
    $\displaystyle a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$.

    [tex]\sum\limits_{n = 0}^\infty {ar^n } [/tex] gives $\displaystyle \sum\limits_{n = 0}^\infty {ar^n } $
    Are you sure $\displaystyle a_n$ can't be written as $\displaystyle a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    Quote Originally Posted by MathIsOhSoHard View Post
    Are you sure $\displaystyle a_n$ can't be written as $\displaystyle a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$
    YES

    It is $\displaystyle a_N=(-1)^N\cdot \left ( \frac{1}{N}+\frac{1}{N-1}\right )$

    You made a sign error.
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    Quote Originally Posted by Plato View Post
    YES

    It is $\displaystyle a_N=(-1)^N\cdot \left ( \frac{1}{N}+\frac{1}{N-1}\right )$

    You made a sign error.
    How do you rewrite $\displaystyle (-1)^{N-1}$ into $\displaystyle (-1)^N$?
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    Quote Originally Posted by mathisohsohard View Post
    how do you rewrite $\displaystyle (-1)^{n-1}$ into $\displaystyle (-1)^n$?
    $\displaystyle (-1)^{n-1}=(-1)^n(-1)^{-1}=-(-1)^n$

    You see $\displaystyle (-1)^{-1}=\frac{1}{-1}=-1$
    Thanks from MathIsOhSoHard
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    Great! Now I get it

    So by using the alternating series test, it can be shown that the series is convergent.

    And to test for absolute convergence, using the comparison test:
    $\displaystyle \left | a_n \right | \le \left | b_n \right |$
    $\displaystyle \sum^\infty_{n=1}\frac{1}{n}\le \sum^\infty_{n=1}(1)^n \left ( \frac{1}{n}+\frac{1}{n-1} \right )$

    Since $\displaystyle \sum^\infty_{n=1}\frac{1}{n}$ is divergent, then the absolute of $\displaystyle b_n$ is divergent too, thus it is not absolutely convergent.

    So conclusion would be that it's conditionally convergent?
    Would the absolute value of the fraction be $\displaystyle \frac{1}{n-1}$ or $\displaystyle \frac{1}{n+1}$? I wasn't quite sure about that.
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    Re: Determining whether a sequence of partial sums is convergent or divergent

    Quote Originally Posted by MathIsOhSoHard View Post
    Great! Now I get it

    So by using the alternating series test, it can be shown that the series is convergent.

    And to test for absolute convergence, using the comparison test:
    $\displaystyle \left | a_n \right | \le \left | b_n \right |$
    $\displaystyle \sum^\infty_{n=1}\frac{1}{n}\le \sum^\infty_{n=1}(1)^n \left ( \frac{1}{n}+\frac{1}{n-1} \right )$

    Since $\displaystyle \sum^\infty_{n=1}\frac{1}{n}$ is divergent, then the absolute of $\displaystyle b_n$ is divergent too, thus it is not absolutely convergent.

    So conclusion would be that it's conditionally convergent?
    Would the absolute value of the fraction be $\displaystyle \frac{1}{n-1}$ or $\displaystyle \frac{1}{n+1}$? I wasn't quite sure about that.
    Very good.
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