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Math Help - series convergence or divergence

  1. #1
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    series convergence or divergence

    Does this series diverge or converge:

    1 + [1*2]/[1*3] + [1*2*3]/[1*3*5] + [1*2*3*4]/[1*3*5*7} + ...
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  2. #2
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    Quote Originally Posted by WartonMorton View Post
    Does this series diverge or converge:

    1 + [1*2]/[1*3] + [1*2*3]/[1*3*5] + [1*2*3*4]/[1*3*5*7} + ...
    This is \sum_{n=1}^\infty a_n, where a_1=1, and a_n=a_{n-1}\cdot\tfrac{n}{2n-1}, and, yes, this series converges.
    For consider that \lim_{n\to\infty}\frac{n}{2n-1}=\frac{1}{2}<1
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  3. #3
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    Quote Originally Posted by WartonMorton View Post
    Does this series diverge or converge:

    1 + [1*2]/[1*3] + [1*2*3]/[1*3*5] + [1*2*3*4]/[1*3*5*7} + ...
    a_{n+1}=a_n \frac{n+1}{2n+1}

    so:

    \frac{a_{n+1}}{a_n}=\frac{1}{2-\frac{1}{n+1}}

    CB
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  4. #4
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    Quote Originally Posted by WartonMorton View Post
    Does this series diverge or converge:

    1 + [1*2]/[1*3] + [1*2*3]/[1*3*5] + [1*2*3*4]/[1*3*5*7} + ...
    1 + \frac{1 \cdot 2}{1 \cdot 3} + \frac{1 \cdot 2 \cdot 3}{1 \cdot 3 \cdot 5} + \frac{1 \cdot 2 \cdot 3 \cdot 4}{1 \cdot 3 \cdot 5 \cdot 7} + \dots

     = \sum_{k = 1}^{\infty}{\frac{k!}{\frac{(2k)!}{k!2^k}}}

     = \sum_{k = 1}^{\infty}\frac{(k!)^22^k}{(2k)!}


    Using the ratio test:

    \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{[(n + 1)!]^22^{n + 1}}{[2(n + 1)]!}}{\frac{(n!)^22^n}{(2n)!}}\right|

     = \lim_{n \to \infty}\left|\frac{(2n)![(n + 1)!]^22^{n + 1}}{[2(n + 1)]!(n!)^22^n}\right|

     = \lim_{n \to \infty}\left|\frac{2(2n)!(n + 1)^2(n!)^22^n}{(2n + 2)(2n + 1)(2n)!(n!)^22^n}\right|

     = \lim_{n \to \infty}\left|\frac{2(n + 1)^2}{2(n + 1)(2n + 1)}\right|

     = \lim_{n \to \infty}\left|\frac{n + 1}{2n + 1}\right|

     = \lim_{n \to \infty}\left(\frac{n + 1}{2n + 1}\right)

     = \lim_{n \to \infty}\frac{1}{2} by L'Hospital's Rule

     = \frac{1}{2}

     < 1.


    So the series is convergent.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Prove It View Post
    1 + \frac{1 \cdot 2}{1 \cdot 3} + \frac{1 \cdot 2 \cdot 3}{1 \cdot 3 \cdot 5} + \frac{1 \cdot 2 \cdot 3 \cdot 4}{1 \cdot 3 \cdot 5 \cdot 7} + \dots

     = \sum_{k = 1}^{\infty}{\frac{k!}{\frac{(2k)!}{k!2^k}}}

     = \sum_{k = 1}^{\infty}\frac{(k!)^22^k}{(2k)!}


    Using the ratio test:

    \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{[(n + 1)!]^22^{n + 1}}{[2(n + 1)]!}}{\frac{(n!)^22^n}{(2n)!}}\right|

     = \lim_{n \to \infty}\left|\frac{(2n)![(n + 1)!]^22^{n + 1}}{[2(n + 1)]!(n!)^22^n}\right|

     = \lim_{n \to \infty}\left|\frac{2(2n)!(n + 1)^2(n!)^22^n}{(2n + 2)(2n + 1)(2n)!(n!)^22^n}\right|

     = \lim_{n \to \infty}\left|\frac{2(n + 1)^2}{2(n + 1)(2n + 1)}\right|

     = \lim_{n \to \infty}\left|\frac{n + 1}{2n + 1}\right|

     = \lim_{n \to \infty}\left(\frac{n + 1}{2n + 1}\right)

     = \lim_{n \to \infty}\frac{1}{2} by L'Hospital's Rule

     = \frac{1}{2}

     < 1.


    So the series is convergent.
    This looks like very hard work compared to the two previous posts, I expect that explains why it took 11 minutes longer to write.

    CB
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