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Math Help - another series...not fibonacci

  1. #1
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    another series...not fibonacci

    Hey Soroban, here's one you may like. Or anyone else, by all means.


    Find the sum of the infinite series:

    \sum_{n=2}^{\infty}\frac{1}{n^{2}+3n-4}
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    Quote Originally Posted by galactus View Post
    Hey Soroban, here's one you may like. Or anyone else, by all means.


    Find the sum of the infinite series:

    \sum_{n=2}^{\infty}\frac{1}{n^{2}+3n-4}

    \sum_{n = 2}^{ \infty} \frac {1}{(n + 4)(n - 1)}

    splitting the last fraction by partial fractions, we find that the sum is actually:

    \sum_{n = 2}^{ \infty} \left( \frac { \frac {1}{5}}{n - 1} - \frac { \frac {1}{5}}{n + 4} \right)

    Thus, we have:

    \frac {1}{5} \sum_{n = 2}^{ \infty} \left( \frac {1}{n - 1} - \frac {1}{n + 4} \right) = \frac {1}{5} \left( 1 - \frac {1}{6} \right.  \left.  + \frac {1}{2} - \frac {1}{7} + \frac {1}{3} - \frac {1}{8} + \frac {1}{4} - \frac {1}{9} + \frac {1}{5} - \frac {1}{10} + \frac {1}{6} - \frac {1}{11} ... \right)

    After writing out a few terms, it does not take long to realize that the - \frac {1}{n + 4} will cancel all the terms from \frac {1}{n - 1} that are \geq \frac {1}{6}

    Thus we can see that:

    \frac {1}{5} \sum_{n = 2}^{ \infty} \left( \frac {1}{n - 1} - \frac {1}{n + 4} \right) = \frac {1}{5} \left( 1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4} + \frac {1}{5} \right) = \frac {137}{300}

    Was that the way you did it galactus?
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    Yes, that's exactly how I approached it. It wasn't too bad. Just fun.

    I like these converging infinite series. Sometimes they're challenging and result in a cool sum.
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    Quote Originally Posted by galactus View Post
    I like these converging infinite series. Sometimes they're challenging and result in a cool sum.
    yeah, i like the ones that end up having pi in the answer

    in general i'm not good with series though
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  5. #5
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    Here's something you may find interesting then.

    You are probably familiar with Euler's Basel problem. You know

    \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}  }{6}

    \sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{{\pi}^{4}  }{90}

    \sum_{n=1}^{\infty}\frac{1}{n^{6}}=\frac{{\pi}^{6}  }{945}

    and on and on.


    Well, Euler found a general formula for the sum of the reciprocals of the EVEN powers.

    \sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\frac{(-1)^{k-1}B_{2k}(2\pi)^{2k}}{2(2k)!}

    So if you wanted \sum_{n=1}^{\infty}\frac{1}{n^{20}}

    You could use the formula and get \frac{174611{\pi}^{20}}{1531329465290625}

    The B_{2k} is the even Bernoulli numbers. They are tied in here also. Isn't it cool how this stuff ties together.

    B_{2}=\frac{1}{6}, \;\ B_{4}=\frac{-1}{30}, \;\ B_{6}=\frac{1}{42}, etc, \;\

    The Bernoulli numbers are an interesting thing to research if you're unfamiliar with them. Just a thought.
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    Quote Originally Posted by galactus View Post
    Here's something you may find interesting then.

    You are probably familiar with Euler's Basel problem. You know

    \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}  }{6}
    I took an advanced calculus class last semester and the professor actually derived this.


    Well, Euler found a general formula for the sum of the reciprocals of the EVEN powers.

    \sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\frac{(-1)^{k-1}B_{2k}(2\pi)^{2k}}{2(2k)!}
    i think he mentioned this formula as well

    The Bernoulli numbers are an interesting thing to research if you're unfamiliar with them. Just a thought.
    I plan on looking up Bernoulli numbers one day (i'm one of those procrastinators, so chances are i won't look them up today even if i said i would). I don't know anything about them other than the fact they are used in this formula.
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  7. #7
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    There's a bunch of different methods of deriving the Basel problem. The most popular using the Taylor expansion for sine. Is that the one your professor used?.
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    Quote Originally Posted by galactus View Post
    There's a bunch of different methods of deriving the Basel problem. The most popular using the Taylor expansion for sine. Is that the one your professor used?.
    yes. i believe he used the Taylor series expansion for \sin ( \pi x )
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    I can derive this with Complex Analysis if anyone is interested.

    Let me just post a real cool infinite summation:
    \sum_{n=1}^{\infty} \frac{1}{n^2+a^2} = \frac{1}{2a}\mbox{coth} \pi a - \frac{1}{2a^2} \mbox{ for }0<a<1.
    -----------------------------
    Leonard Euler's immortal approach was not formal. His idea was that a function can be factored by its zero's like a polynomial. Though his idea was informal it attracted attention. The general theorem is known as the Weierstrauss Factorization Theorem.

    Off Topic: How is "Weierstrauss" pronounced properly, without an Americanized accent. Because I hated how my professor pronounced his name, it is so....so Americanized, if you know what I mean.
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