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Math Help - Converge or Diverge?

  1. #1
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    Converge or Diverge?

    a_k=\frac{4^k(k!)^2}{(2k)!}

    Does \sum_1^\infty (-1)^ka_k converge absolutely, converge conditionally, or diverge?

    Any help on this? Not really sure where to even begin...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by paupsers View Post
    a_k=\frac{4^k(k!)^2}{(2k)!}

    Does \sum_1^\infty (-1)^ka_k converge absolutely, converge conditionally, or diverge?

    Any help on this? Not really sure where to even begin...
    absolute convergence means the series \sum |a_k| converges. Conditional convergence means \sum (-1)^ka_k converges, but \sum a_k doesn't.

    Here, I would use the ratio test. so try that.
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  3. #3
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    if \sum a_n<\infty, and obviously here a_n\ge0, then \sum(-1)^n a_n<\infty, and this is obvious since \sum |(-1)^n a_n|=\sum a_n<\infty.
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  4. #4
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    Quote Originally Posted by paupsers View Post
    a_k=\frac{4^k(k!)^2}{(2k)!}

    Does \sum_1^\infty (-1)^ka_k converge absolutely, converge conditionally, or diverge?

    Any help on this? Not really sure where to even begin...
    I can tell you without computation that a_k\geq 1 for all k, which pretty much settles it...

    So, why? Because \frac{1}{a_k}=\frac{1}{2^{2k}}{2k\choose k} is the probability for a symmetric simple random walk on the integers to be back at its starting point at time 2k (the binomial coefficient comes from choosing which k steps go to the right); hence \frac{1}{a_k}\leq 1. Or just because {2k\choose k} is the number of subsets of \{1,\ldots, 2k\} with k elements, which is less than 2^{2k}, the number of subsets of \{1,\ldots,2k\}.

    By the way, Stirling estimate gives a_n\sim\sqrt{\pi n} if I remember well, hence the ratio and root tests would be unconclusive.
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