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Math Help - Convergence of Series

  1. #1
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    Convergence of Series

    Good evening All,

    I am wrapping up my final problem in the Root & Ratio Test section. I have the following:

    \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)}  =  \sum_{n=1}^{\infty}\frac{2n-1}{3n-2}

    So...

    \displaystyle\lim_{n\rightarrow\infty}|\frac{2(n+1  )-1}{3(n+1)-1}*\frac{3n-2}{2n-1}|

    So...

    \displaystyle\lim_{n\rightarrow\infty}|\frac{2n+1}  {3n+2}*\frac{3n-2}{2n-1}|

    So...

    \displaystyle\lim_{n\rightarrow\infty}|\frac{6n^2-n-2}{6n^2+n-2}|=1

    Inconclusive by the Ratio Test...

    What's the next plan?

    If I apply the Divergence Test (which should probably have been the first thing I did), I find...

    \displaystyle\lim_{n\rightarrow\infty}\frac{2n-1}{3n-2}  =  \frac{2}{3}  \neq  0

    Therefore, by the Divergence Test the series Diverges

    Does anyone agree or disagree with this?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    The first line of yours is not true at all!
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  3. #3
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    \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} = \sum_{n=1}^{\infty}\frac{2n-1}{3n-2}


    Could you please explain? I thought that the two meant essentially the same thing.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Compute to n=1,2,3,4 instead to inf, in both sides...
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  5. #5
    MHF Contributor chisigma's Avatar
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    If we write the series as...

    \displaystyle \sum_{n=1}^{\infty} \frac{1\ 3\ \ 5\ \dots \ (2n-1)}{1\ 4\ 7\ \dots \ (3n-2)} = \sum_{n=1}^{\infty} a_{n} (1)

    ... the DE defining the a_{n} is...

    \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1 (2)

    From (2) it follows that is...

    \displaystyle \lim_{n \rightarrow \infty} a_{n} =0

    \displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}} = \frac{2}{3} < 1 (3)

    ... so that the series (1) converges...

    Kind regards

    \chi \sigma
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  6. #6
    Senior Member yeKciM's Avatar
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    Cauchy convergence criteria:

    \displaystyle \sum_{n=1}^{\infty}a_n

    \displaystyle q= \lim_{n\to\infty} \sqrt[n]{a_n}

    \displaystyle q=\begin{cases}<br />
 q<1& converges  \\ <br />
 q>1& diverges  \\ <br />
 q=1 & ? need \;more\; testing  <br />
\end{cases}


    D'alambert convergence criteria :

    \displaystyle \sum_{n=1}^{\infty}a_n

    \displaystyle q= \lim_{n\to\infty} \frac{a_{n+1}}{a_n}

    \displaystyle q=\begin{cases}<br />
 q<1& converges  \\ <br />
 q>1& diverges  \\ <br />
 q=1 & ? need\; more\; testing  <br />
\end{cases}


    Raabel's convergence criteria:

    \displaystyle \sum_{n=1}^{\infty}a_n

    \displaystyle q= \lim_{n\to\infty} n[ \frac{a_{n+1}}{a_n} -1]


    \displaystyle q=\begin{cases}<br />
 q>1& converges  \\ <br />
 q<1& diverges  \\ <br />
 q=1 & ? need\; more\; testing  <br />
\end{cases}


    for alternative series u'll use
    Leibnitz convergence criteria:

    \displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}a_n

    converges if the conditions are met that :


    1 \displaystyle \lim_{n\to\infty} a_n = 0

    2 a_n monotonically decreasing

    if not then it diverges


    or u can test it by the definition that say that \displaystyle \sum_{n=1}^{\infty}a_n converges if \displaystyle S_N= \sum_{n=1}^{\infty}a_n=a_1+a_2+ . . . + a_N sequence (S_N)_{N\in\mathbb{N}} which is sequence of partial sum, is :

    \displaystyle \lim_{N\to\infty}S_N =\begin{cases}<br />
 final \;number\Rightarrow  \;series \;converges  \\ <br />
 infinite\; or\; does \;not \;exist\Rightarrow  \;series\; diverges   <br />
\end{cases}


    there are more criteria but, it's not for real numerical series... it's for function series ...
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  7. #7
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    Quote Originally Posted by chisigma View Post
    If we write the series as...

    \displaystyle \sum_{n=1}^{\infty} \frac{1\ 3\ \ 5\ \dots \ (2n-1)}{1\ 4\ 7\ \dots \ (3n-2)} = \sum_{n=1}^{\infty} a_{n} (1)

    ... the DE defining the a_{n} is...

    \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1 (2)

    From (2) it follows that is...

    \displaystyle \lim_{n \rightarrow \infty} a_{n} =0

    I'm completely lost. We have a_n=1

    \displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}} = \frac{2}{3} < 1 (3)

    ... so that the series (1) converges...

    Kind regards

    \chi \sigma
    Hello,

    How did we determine that...

    \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1

    I'm completely lost. We have a_{n+1}=1 and a_n=0.

    Yet we have \displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1  }}{a_n}=\frac{2}{3}

    Wouldn't we have \displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1  }}{a_n}  =  \lim_{n\rightarrow\infty}\frac{1}{0} = undefined ?

    And what does "DE" mean?

    Sorry for all the questions, just trying to fully understand this.
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  8. #8
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by MechEng View Post
    Hello,

    How did we determine that...

    \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1

    I'm completely lost. We have a_{n+1}=1 and a_n=0.

    Yet we have \displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1  }}{a_n}=\frac{2}{3}

    Wouldn't we have \displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1  }}{a_n}  =  \lim_{n\rightarrow\infty}\frac{1}{0} = undefined ?

    And what does "DE" mean?

    Sorry for all the questions, just trying to fully understand this.


    \displaystyle \sum_{n=1}^{\infty} \frac{1\ 3\ \ 5\ \dots \ (2n-1)}{1\ 4\ 7\ \dots \ (3n-2)} = \sum_{n=1}^{\infty} a_{n}

    if u look at that all under sum as  a_n

    then your a_{n+1} is:

    \displaystyle \sum_{n=1}^{\infty} \frac{1\ 3\ \ 5\ \dots \ (2n-1)(2n+1)}{1\ 4\ 7\ \dots \ (3n-2)(3n+1)} = \sum_{n=1}^{\infty} a_{n+1}

    "for n+1 is going to be [2(n+1)-1]=(2n+1) and [3(n+1)-2]=(3n+1) "

    so when u put it in some of those criteria of convergence those a lot of members would be reduced...

    like in D'alambert convergence criteria:

    \displaystyle q=\lim_{n\to\infty} \frac {a_{n+1}}{a_n}=\lim_{n\to\infty} \frac {\frac{1\ 3\ \ 5\ \dots \ (2n-1)(2n+1)}{1\ 4\ 7\ \dots \ (3n-2)(3n+1)}}{\frac{1\ 3\ \ 5\ \dots \ (2n-1)}{1\ 4\ 7\ \dots \ (3n-2)}}=\lim_{n\to\infty}\frac{2n+1}{3n+1}=\frac {2}{3}<1

    so it converges okay ?


    Edit: Sorry i corrected that typo
    Last edited by yeKciM; August 5th 2010 at 07:56 AM.
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  9. #9
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    Ah, yes... I follow that. Thank you!

    Quick question...

    \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} = \sum_{n=1}^{\infty}\frac{(2n-1)!}{(3n-2)!}

    Does make my statement any better?
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  10. #10
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by MechEng View Post
    Ah, yes... I follow that. Thank you!

    Quick question...

    \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} = \sum_{n=1}^{\infty}\frac{(2n-1)!}{(3n-2)!}
    Does make my statement any better?
    no,

    (2n-1)!=(2n-1)(2n-2)(2n-3)... (2n-m)!



    so it's not

    Edit: edited 2 times first time correct to incorrect and now back to correct (now i'm confused sorry )


    maybe just maybe should be :

    \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5...(2n-1)}{1*4*7...(3n-2)} =\sum_{n=1}^{\infty}\frac{\displaystyle \prod_{n=1}^{\infty}(2n-1)}{\displaystyle \prod_{n=1}^{\infty}(3n-2)}


    Edit: didn't know how to write product
    Last edited by yeKciM; August 5th 2010 at 09:25 AM.
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  11. #11
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by MechEng View Post
    Hello,

    How did we determine that...

    \displaystyle a_{n+1} = a_{n}\ \frac{2n+1}{3n+1} , a_{1}=1

    ... and what does "DE" mean?...
    Is...

    \displaystyle a_{n} = \frac{1\ 3\ 5\ \dots (2n-1)}{1\ 5\ 7\ \dots \ (3n-2)} (1)

    ... and...

    \displaystyle a_{n+1} = \frac{1\ 3\ 5\ \dots (2n-1)\ (2n+1)}{1\ 5\ 7\ \dots \ (3n-2)\ (3n+1)} (2)

    ... so that from (1) and (2) You obtain...

    \displaystyle \frac{a_{n+1}}{a_{n}} = \frac{2n+1}{3n+1} (3)

    The term 'DE' means difference equation...

    Kind regards

    \chi \sigma
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  12. #12
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    Quote Originally Posted by yeKciM View Post
    no,

    (2n-1)!=(2n-1)(2n-2)(2n-3)... (2n-m)!

    for n=1 it would be :

     1\cdot 0\cdot ( -1)....
    But it's (2n-1)!

    So...

    (2(1)-1)+(2(2)-1)+(2(3)-1)+(2(4)-1)+...+(2(n)-1)

    Or...

    1+3+5+7+...+(2n-1)

    The n is what's changing.

    Right?
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  13. #13
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by MechEng View Post
    But it's (2n-1)!

    So...

    (2(1)-1)+(2(2)-1)+(2(3)-1)+(2(4)-1)+...+(2(n)-1)

    Or...

    1+3+5+7+...+(2n-1)

    The n is what's changing.

    Right?
    hmmmm...
    first there are no "+" there for Factorial

     5!=5\cdot4\cdot3\cdot2\cdot1

    1\cdot3\cdot5\cdot. . .\cdot(2n-1)=\displaystyle \prod_{n=1} ^{\infty} (2n-1)


    but if yours is true then for let's say n=2 is going to be

     n=2

     (2n-1)! = 3! = 3\cdot2\cdot1

    and that's not what u have on left side up there somewhere in starting sum


    P.S. I corrected that post (#10)
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  14. #14
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    Ah, yes...

    Not sure how that happened.
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  15. #15
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    Ok, I think I've gotten way too far off track here.

    Starting with the original series:

    \displaystyle\sum_{n=1}^{\infty}\frac{1*3*5*...*(2  n-1)}{1*4*7*...*(3n-2)}

    What should my \displaystyle a_n be equal to?

    I understand all of the tests, but cannot seem to recognize the original series as something that I can work with.

    I'm sure you've pointed out how to go from the original series to a more useable form, but I am having a hard time following you.
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