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Math Help - Absolute or conditional convergence question

  1. #1
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    Absolute or conditional convergence question

    how do i find wheter this is absolute or conditional convergence, can someone offer some guide & advice,thank you so much. im out of ideas for this.

    i) \sum^\infty _{n=1} \frac {sin(\frac {n\pi}{2})}{n!}
    ii) \sum^\infty _{n=1} (-1)^n \frac {n}{ln n}

    thank you.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    accidentally posted twice
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  3. #3
    MHF Contributor Calculus26's Avatar
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    always check for absolute convergence first because if a series converges absolutely you're done. If not then check for conditional convergence.

    For 1) sin(npi/2) = -1,0 or1

    |sin(npi/2|/n! < 1/n! and the series for 1/n! converges (easy to check with ratio test)

    lim n/ln(n) = inf therefore the series does not converge absolutely

    also since lim n/ln(n) = inf then series (-1)^n n/ln(n) diverges

    as lim (-1)^n n/ln(n) = DNE
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    thank you so much for replying Calculus26.from your explanation, both equations converge conditionally, am i right?

    so for sin(npi/2) = -1,0 or1, the series converges conditionally?
    for [tex]

    please help to clarify, i still can't understand fully. also, what does DNE mean?

    sorry to trouble you, thank you so much for helping.
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  5. #5
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    I think that the second must be like that: \sum_{i=2}^{\infty}. To that serie, you can see that \lim_{n\to\infty}\dfrac{n}{\ln(n)}\neq 0 (easy to see using L'H˘pital), then, doesn't converges.
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    for <br />
i)<br />
\frac {sin(\frac {n\pi}{2})}{n!}<br />
    |sin(npi/2|/n! < 1/n & 1/n! converges,

    the equation converge absolutely.

    ii)lim n/ln(n) = inf then series (-1)^n n/ln(n) diverges

    the equation converge conditionally.

    is this correct? im stuck for proving with question 2, need someone to help clarify,thank you so much.
    Last edited by anderson; February 5th 2010 at 06:52 AM.
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  7. #7
    MHF Contributor Calculus26's Avatar
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    you have it correct for 1) converges abs

    However 2) diverges period since
    also since lim n/ln(n) = inf then series (-1)^n n/ln(n) diverges

    as lim (-1)^n n/ln(n) = DNE
    Remeber if the sequence does not converge to 0 the series diverges.

    there are 3 possibilities -- a series can conv abs,con conditionally, or diverges

    An eg of conditional convergence is the series (-1)^n/ n

    The series |(-1)^n/ n| = series 1/n diverges

    however series (-1)^n/ n converges by the alternating series test

    lim1/n = 0
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    just confirmed, there is an error in question 2,n=2,just like the feedback from felper, thank you felper.

    ii) \sum^\infty _{n=2} (-1)^n \frac {n}{ln n}

    how do i prove this equation converge now,absolute or conditional., can anyone help me. thank you so much friends for helping me thus far, really appreciate.


    thank you.
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  9. #9
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    that was already answered above.

    besides that series diverges, so there's no chance of absolute convergence nor conditional convergence.
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  10. #10
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    thank you everyone for helping, but im really confused as to why this series diverges and why <br /> <br />
\lim_{n\to\infty}\dfrac{n}{\ln(n)}\neq 0<br />
.

    if someone can explain this it will be very helpful. thank you so much everyone for helping, sorry to trouble all of you, im really out of ideas.

    from the comments:
    An eg of conditional convergence is the series (-1)^n/ n

    The series |(-1)^n/ n| = series 1/n diverges

    however series (-1)^n/ n converges by the alternating series test

    lim1/n = 0

    is it that (-1)^n/ n is taken from the original equation, just minus the ln from the equation? just wondering how we got that.

    just wonderring about this few explanations, if someone could help me, it will be great. thank you so much.
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  11. #11
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    Given any series what so ever, \sum {a_n } , if it happens that  \left( {a_n } \right) \not\to 0 then the series diverges.
    That is known as the first divergent test.
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    thank you everyone for helping,want to know how we conclude that
    <br /> <br />
\lim_{n\to\infty}\dfrac{n}{\ln(n)}\neq 0<br />
.

    using L Hospital rule.so far i only can derive this

    \frac {n/n}{(lnn)/n}
    = \frac {1}{\frac{ln n}{n}}

    is there any method for this,thank you for helping
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  13. #13
    ynj
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    Quote Originally Posted by anderson View Post
    thank you everyone for helping,want to know how we conclude that
    <br /> <br />
\lim_{n\to\infty}\dfrac{n}{\ln(n)}\neq 0<br />
.

    using L Hospital rule.so far i only can derive this

    \frac {n/n}{(lnn)/n}
    = \frac {1}{\frac{ln n}{n}}

    is there any method for this,thank you for helping
    you can prove \lim_{x\rightarrow\infty}\frac{\ln x}{x}=0by using L'Hospital rule. Then by Heine Principle you can derive what you want.
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  14. #14
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    i think i get it now, so this means that the limit n/ln(n) equals to infinity.

    is it so? thank you for helping me to understand.

    thank you for all your help & guidance.
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