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Math Help - prove that these series converge to values given

  1. #1
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    prove that these series converge to values given

    I know how to determine what the series \sum_{n=1}^{\infty}n^pr^{n-1} converge to (for -1 < r < 1) for positive integral values of p:

    p = 1: \frac{1}{(1 - r)^2}

    p = 2: \frac{1+r}{(1 - r)^3}

    p = 3: \frac{1 + 4r + r^2}{(1 - r)^4}

    p = 4: \frac{1 + 11r + 11r^2 + r^3}{(1 - r)^5}

    p = 5: \frac{1 + 26r + 66r^2 + 26r^3 + r^4}{(1 - r)^6}

    The general pattern is:

    \frac{a_0 + a_1r + ... + a_{p-2}r^{p-2} + a_{p-1}r^{p-1}}{(1 - r)^{p+1}}

    where the coefficients a_0, a_1, ... a_{p-1} are given by the pth row of Euler's triangle (there is an article on the triangle at www.mathworld.com).

    So I figured this out but I don't know how to prove it formally. Any suggestions?
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  2. #2
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    Quote Originally Posted by icemanfan View Post
    I know how to determine what the series \sum_{n=1}^{\infty}n^pr^{n-1} converge to for positive integral values of p:

    p = 1: \frac{1}{(1 - r)^2}

    p = 2: \frac{1+r}{(1 - r)^3}

    p = 3: \frac{1 + 4r + r^2}{(1 - r)^4}

    p = 4: \frac{1 + 11r + 11r^2 + r^3}{(1 - r)^5}

    p = 5: \frac{1 + 26r + 66r^2 + 26r^3 + r^4}{(1 - r)^6}

    The general pattern is:

    \frac{a_0 + a_1r + ... + a_{p-2}r^{p-2} + a_{p-1}r^{p-1}}{(1 - r)^{p+1}}

    where the coefficients a_0, a_1, ... a_{p-1} are given by the pth row of Euler's triangle (there is an article on the triangle at www.mathworld.com).

    So I figured this out but I don't know how to prove it formally. Any suggestions?
    Here is an idea

    First re-index the series to start at zero. this gives

    \displaystyle \sum_{n=0}^{\infty}(n+1)^pr^n

    Now using the binomial theorem of the factor

    \displaystyle (n+1)^p=\sum_{k=0}^{p}\binom{p}{k}n^{k}

    Combing the two above gives

    \displaystyle \sum_{n=0}^{\infty}\sum_{k=0}^{p}\binom{p}{k}n^{k}  r^n

    Now lets analyze the pattern of

    \displaystyle k=0; \quad r^n
    \displaystyle k=1; \quad n^{1}r^n=r \frac{d}{dr}r^n
    \displaystyle k=2; \quad n^{2}r^n=r \frac{d}{dr}r \frac{d}{dr}r^n
    \displaystyle k=p; \quad n^pr^n =\left( r\frac{d}{dr}\right)^pr^n

    \displaystyle \sum_{n=0}^{\infty}\sum_{k=0}^{p}\binom{p}{k}n^{k}  r^n=\sum_{n=0}^{\infty}\sum_{k=0}^{p}\binom{p}{k}\  left( r\frac{d}{dr}\right)^kr^n=\sum_{k=0}^{p}\binom{p}{  k}\left( r\frac{d}{dr}\right)^k\sum_{n=0}^{\infty}r^n=\sum_  {k=0}^{p}\binom{p}{k}\left( r\frac{d}{dr}\right)^k\frac{1}{1-r}

    Note that the derivatives can be pulled out of the infinite sum for |r|<1 because the geometric series is uniformly convergent.
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