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Math Help - Prove some identities!

  1. #31
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    Quote Originally Posted by Drexel28 View Post


    Here is quite a beautiful one (in my opinion).

    Problem: Let \Omega\subseteq\mathbb{R}^n be the set of all (x_1,\cdots,x_n) such that x_j\geqslant 0,\text{ }j\in[n] and \displaystyle \sum_{j=1}^{n}x_j=1. Compute


    \displaystyle \int\cdots\int_{\Omega}x^{\alpha_1-1}\cdots x^{\alpha_n-1}\text{ }dx_1\cdots dx_n


    where \text{Re }\alpha_k>0,\text{ }k\in[n].
    the integrand should be x_1^{\alpha_1-1} \cdots x_n^{\alpha_n - 1}.

    call the integral I_n and substitute x_n=y_n and x_i = (1-y_n)y_i for i < n. then the Jacobian would be (1-y_n)^{n-1} and thus

    \displaystyle I_n=I_{n-1}\int_0^1 y_n^{\alpha_n -1}(1-y_n)^{\alpha_1 + \cdots + \alpha_{n-1}} dy_n = I_{n-1} B(\alpha_n, \alpha_1 + \cdots + \alpha_{n-1}+1),

    where B(-,-), as usual, is the beta function. it's easy now to see that

    \displaystyle I_n = \frac{\Gamma(\alpha_1) \Gamma(\alpha_2) \cdots \Gamma(\alpha_n)}{\Gamma(\alpha_1 + \cdots + \alpha_n +1)}.
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  2. #32
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Drexel28 View Post
    \displaystyle \sum_{j=1}^{n}j{n\choose j}
    Consider the more general: \sum_{k=0}^n {\binom{n}{k}\cdot \binom{k}{j}}

    First note that what we are doing is choosing subsets of size j by the following procedure:

    Fix k\in\{0,..,n\}

    1. Choose a subset of size k .

    2. Now choose a subset of size j from the subset chosen in 1 - if possible.

    It is clear that we are overcounting, but by how much? Well, any subset of size j is part of \binom{n-j}{k-j} subsets of size k - that's the number of ways of completing the set -, that is \binom{n}{k}\cdot \binom{k}{j} = \binom{n-j}{k-j}\cdot \binom{n}{j}

    So in fact we have \sum_{k\geq 0}{\binom{n}{k}\cdot \binom{k}{j}} = \sum_{k\geq 0}{\binom{n}{j}\cdot \binom{n-j}{k-j}} = \binom{n}{j}\cdot 2^{n-j}

    Now plug j = 1 and we get \sum_{k=0}^{n}{\binom{n}{k}\cdot k} = n\cdot 2^{n-1}


    EDIT: I am going to add a couple of problems for you to enjoy.

    1. e = \prod_{k = 1}^{\infty}{\left(\frac{2^k}{2^k-1}\right)^{\frac{\phi(k)}{k}}} where \phi is Euler's Totien function.

    2. This one is beautiful too, let \pi \in S_n (*) we define \text{inv}(\pi) to be the number of pairs of integers (i,j) that satisfy 1\leq i < j \leq n and \pi(j) < \pi ( i) - such a pair is called an inversion.

    Show that: \sum_{\pi \in S_n}{q^{\text{inv}(\pi)}}=(1+q)\cdot ... \cdot (1+q+...+q^{n-1})

    What is the average number of inversions ?

    (*) See here
    Last edited by PaulRS; December 24th 2010 at 05:58 AM.
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  3. #33
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by PaulRS View Post

    2. This one is beautiful too, let \pi \in S_n (*) we define \text{inv}(\pi) to be the number of pairs of integers (i,j) that satisfy 1\leq i < j \leq n and \pi(j) < \pi ( i) - such a pair is called an inversion.

    Show that: \sum_{\pi \in S_n}{q^{\text{inv}(\pi)}}=(1+q)\cdot ... \cdot (1+q+...+q^{n-1})
    See if this suffices:

    Spoiler:

    This is fairly easy if you know the (common?) combinatorial fact that that the number of j-inversions of [n] is the (j,n)^{\text{th}} Mahnonian number which is the coefficient of x^j in \displaystyle \prod_{k=1}^{n}\sum_{r=0}^{k-1}x^r denotes this by \displaystyle \text{coeff}_j\left(\prod_{k=1}^{n}\sum_{r=0}^{k-1}x^r\right). Thus, note that for any n^{\text{th}} degree polynomial we have that \displaystyle p(x)=\sum_{k=0}^{n}x^k \text{coeff}_k(p(y)) (clearly). Note though that that the degree of \displaystyle (1+x)\cdots(1+x^{n-1}) is \displaystyle (n-1)+(n-2)+\cdots+1=\frac{n(n-1)}{2}={n\choose 2}. That said it's also clear that a permutation of [n] may have (and indeed has at least one) j-inversion for \displaystyle j=0,\cdots,{n\choose 2}. Thus,


    \displaystyle \begin{aligned}\sum_{\pi\in S_n}q^{\text{inv}(\pi)} &= \sum_{k=0}^{{n\choose 2}}q^k\text{coeff}_k\left(\prod_{k=1}^{n}\sum_{r=0  }^{k-1}x^r\right)\\ &= \prod_{k=1}^{n}\sum_{r=0}^{k-1}q^r\end{aligned}

    as required.


    Let me know if that identity I stated is too obscure.

    Last edited by Drexel28; December 24th 2010 at 09:53 AM.
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  4. #34
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by PaulRS View Post
    1. e = \prod_{k = 1}^{\infty}{\left(\frac{2^k}{2^k-1}\right)^{\frac{\phi(k)}{k}}} where \phi is Euler's Totien function.
    Let F(x) = \sum_{k=1}^\infty \phi(k) \frac{x^k}{1-x^k}.

    We have F(x) = \sum_{k=1}^\infty \sum_{n=1}^\infty \phi(k)x^{nk} = \sum_{m=1}^\infty \sum_{d|m}\phi(d)x^m = \sum_{m=1}^\infty mx^m = x\left(\frac{1}{1-x}\right)^2.

    Now consider

    G(t)=\int_{0}^t\frac{F(x)}{x}dx = \sum_{k=1}^\infty \phi(k) \int_0^t \frac{x^{k-1}}{1-x^k} dx = \sum_{k=1}^\infty \frac{\phi(k)}{k} \int_0^{t^k} \frac{du}{1-u} = -\sum_{k=1}^\infty \frac{\phi(k)}{k} \log(1-t^k).

    We also have G(t) = \int_{0}^t\left(\frac{1}{1-x}\right)^2 dx = \frac{1}{1-t} - 1

    Hence G(1/2) = 1 = -\sum_{k=1}^\infty \frac{\phi(k)}{k} \log(1-2^{-k}) = \sum_{k=1}^\infty \frac{\phi(k)}{k} \log(\frac{2^k}{2^k-1}).

    Taking the exponential, we get the desired equality.
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  5. #35
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    See if this suffices:

    Spoiler:

    This is fairly easy if you know the (common?) combinatorial fact that that the number of j-inversions of [n] is the (j,n)^{\text{th}} Mahnonian number which is the coefficient of x^j in \displaystyle \prod_{k=1}^{n}\sum_{r=0}^{k-1}x^r denotes this by \displaystyle \text{coeff}_j\left(\prod_{k=1}^{n}\sum_{r=0}^{k-1}x^r\right). Thus, note that for any n^{\text{th}} degree polynomial we have that \displaystyle p(x)=\sum_{k=0}^{n}x^k \text{coeff}_k(p(y)) (clearly). Note though that that the degree of \displaystyle (1+x)\cdots(1+x^{n-1}) is \displaystyle (n-1)+(n-2)+\cdots+1=\frac{n(n-1)}{2}={n\choose 2}. That said it's also clear that a permutation of [n] may have (and indeed has at least one) j-inversion for \displaystyle j=0,\cdots,{n\choose 2}.

    \displaystyle \begin{aligned}\sum_{\pi\in S_n}q^{\text{inv}(\pi)} &= \sum_{k=0}^{{n\choose 2}}q^k\text{coeff}_k\left(\prod_{k=1}^{n}\sum_{r=0  }^{k-1}x^r\right)\\ &= \prod_{k=1}^{n}\sum_{r=0}^{k-1}q^r\end{aligned}

    as required.


    Let me know if that identity I stated is too obscure.

    I feel as though it's a little cheap to prove this theorem via the other one (the fact that the number of j-inversions is \text{coeff}_j\left((1+x)\cdots(1+x+\cdots+x^{n-1})\right)) without proving it first. So, I'll modify the proof of that theorem to prove this problem directly.

    Spoiler:


    Claim: \displaystyle \sum_{\pi\in S_n}q^{\text{inv}(\pi)}=\prod_{k=1}^{n}\sum_{r=0}^  {k-1}q^r for n\geqslant 2.
    Proof: We proceed by induction. For n=2 this is clear since S_2=\left\{\text{id}_{[2]},(1,2)\} and so


    \displaystyle \sum_{\pi\in S_2}q^{\text{inv}(\pi)}=q^{\text{inv}(\text{id}_{[2]})}+q^{\text{inv}((1,2))}=q^0+q^1=1+q


    as claimed. Assume now that this is true for n and notice that S_{n+1} is partitioned into two blocks call the blocks B_1 and B_2. Namely, those which fix n+1 and those that don't. Identify now each \pi\in S_{n+1} with it's associated n+1-tuple. Then, S_{n+1} is partitioned into the two blocks of the form (\ast,\cdots,\ast,n+1) and (\ast,\cdots,n+1,\cdots,\ast). Note that if \pi is of the second form with \pi=(\underbrace{\ast,\cdots,\ast}_{\ell},n+1,\cdo  ts,\ast) and if \sigma is the element of S_n which results from removing that \ell+1^{\text{th}} slot (the one containing n+1) then \text{inv}(\pi)=\text{inv}(\sigma)+\ell. Thus, it's clear from this that

    \displaystyle \sum_{\pi\in B_2}q^{\text{inv}(\pi)}=\sum_{\ell=1}^{n}q^\ell\su  m_{\pi\in S_n}q^{\text{inv}(\pi)}


    Noticing then that for each \pi\in B_1 we have that \pi is just a permutation of [n] with the extra condition that n+1 always maps to n+1 (and in particular it adds not inversions) we can see that

    \displaystyle \sum_{\pi\in B_1}q^{\text{inv}(\pi)}=\sum_{\pi\in S_n}q^{\text{inv}(\pi)}

    Thus,


    \displaystyle \begin{aligned}\sum_{\pi\in S_{n+1}}q^{\text{inv}(\pi)} &= \sum_{\pi\in B_1}q^{\text{inv}(\pi)}+\sum_{\pi\in B_2}q^{\text{inv}(\pi)}\\ &=\sum_{\pi\in S_n}q^{\text{inv}(\pi)}+\sum_{\ell=1}^{n}q^{\ell}\  sum_{\pi\in S_n}q^{\text{inv}(\pi)}\\ &= \sum_{\pi\in S_n}q^{\text{inv}(\pi)}\left(1+q+\cdots+q^n)\\ &= \prod_{k=1}^{n}\sum_{r=0}^{k-1}q^r\sum_{r=0}^{n}q^r\\ &= \prod_{k=1}^{n+1}\sum_{r=0}^{k-1}q^r\end{aligned}


    and the induction is complete.

    Last edited by Drexel28; December 24th 2010 at 02:51 PM.
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  6. #36
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I feel as though it's a little cheap to prove this theorem via the other one (the fact that the number of j-inversions is \text{coeff}_j\left((1+x)\cdots(1+x+\cdots+x^{n-1})\right)) without proving it first. So, I'll modify the proof of that theorem to prove this problem directly.
    Hahaha, that's true. When I saw the post above I thought : "But that result you are mentioning is the whole problem itself" but now it's ok ;P
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  7. #37
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    Here are a few sums for you to calculate. However these may be too easy...


    \displaystyle\sum_{n=1}^\infty \frac{\text{H}_n}{2^nn}


    \displaystyle\sum_{n=1}^\infty \frac{\text{H}_n}{n^2}

    and one that involves much more algebra:


    \displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{3n+1}


    Where \text{H}_n is the n'th harmonic number.
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  8. #38
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    Here are a few sums for you to calculate. However these may be too easy...

    \displaystyle\sum_{n=1}^\infty \frac{\text{H}_n}{2^nn}

    Where \text{H}_n is the n'th harmonic number.
    Spoiler:

    First consider  \displaystyle \sum_{n=1}^\infty H_n x^n :

     \displaystyle \sum_{n=1}^\infty H_n x^n = \sum_{n=1}^\infty\sum_{i=1}^n \frac1i x^n = \sum_{i=1}^\infty \frac1i \sum_{n=i}^\infty x^n = \sum_{i=1}^\infty \frac1i \frac{x^i}{1-x} = -\frac{\log(1-x)}{1-x}

    -------

    Now define  \displaystyle F(x) = \sum_{n=1}^\infty \frac{H_n}n x^n . We're after  F(\tfrac12) .

    From the work above, we see  \displaystyle F(x) = -\int_0^x\frac{\log(1-t)}{t(1-t)} dt = -\int_0^x \frac{\log(1-t)}t dt + \frac12 \log^2(1-x) .

    Let the integral equal  G(x) .

    Thus  \displaystyle F(\tfrac12) = G(\tfrac12) + \frac12\log^2 2 = \frac1{12}\pi^2 .

    -------

    We can obtain  G(\tfrac12) by deriving the identity  \displaystyle G(x) + G(1-x) = \frac16\pi^2 - \log(x)\log(1-x) for  0 < x < 1 , which can be shown directly.

    Edit: I see  G(\tfrac12) has already been tackled earlier in this thread!
    Last edited by chiph588@; December 26th 2010 at 03:24 PM.
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  9. #39
    MHF Contributor chiph588@'s Avatar
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    Here's my rebuttal:

    For  \alpha > 1 , evaluate  \displaystyle \lim_{t\to\infty} \frac t{\log t}\sum_{n=0}^\infty \frac1{\alpha^n+t} .
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  10. #40
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    Here are a few sums for you to calculate. However these may be too easy...


    \displaystyle\sum_{n=1}^\infty \frac{\text{H}_n}{2^nn}
    Spoiler:

    Begin by recalling that the Cauchy product of two power series \displaystyle \sum_{n\in\mathbb{N}}a_n z^n and \displaystyle \sum_{n\in\mathbb{N}}b_n z^n is \displaystyle \sum_{n\in\mathbb{N}}\sum_{k=0}^{n}a_k b_{n-k}z^n. So, in particular


    \displaystyle \left(\sum_{n\in\mathbb{N}}a_n z^n\right)\left(\sum_{n\in\mathbb{N}}z^n\right) &= \sum_{n\in\mathbb{N}}\sum_{k=0}^{n}a_k z^k


    for where both series are defined. Thus, note that since \displaystyle -\ln(1-z)=\sum_{n\in\mathbb{N}}\frac{z^n}{n} that the above implies that


    \displaystyle \begin{aligned}\frac{-\ln(1-z)}{1-z} & = \left(\sum_{n\in\mathbb{N}}\frac{z^n}{n}\right)\le  ft(\sum_{n\in\mathbb{N}}z^n\right)\\ &= \sum_{n\in\mathbb{N}}\sum_{k=1}^{n}\frac{1}{k}z^n\  \ &= \sum_{n\in\mathbb{N}}H_n z^n\end{aligned}



    Therefore,


    \displaystyle \begin{aligned}\sum_{n\in\mathbb{N}}H_n \frac{\left(\frac{1}{2}\right)^n}{n} &= \sum_{n\in\mathbb{N}}H_n\int_0^{\frac{1}{2}} z^{n-1}\text{ }dz\\ &= \int_0^{\frac{1}{2}} \sum_{n\in\mathbb{N}}H_n z^{n-1}\text{ }dz\\ &= \int_0^{\frac{1}{2}} \frac{-\ln(1-z)}{z(1-z)}\text{ }dz\end{aligned}


    Thus, it suffices compute this integral. To do this we merely note that


    \displaystyle \begin{aligned}\int_0^{\frac{1}{2}}\frac{-\ln(1-z)}{z(1-z)}\text{ }dz &= -\int_0^1 \frac{\ln(1-\frac{\xi}{2})}{\xi(1-\frac{\xi}{2})}\text{ }d\xi\\ &= -\int_0^1\frac{\ln(1-\frac{\xi}{2})}{\xi}\text{ }d\xi-\frac{1}{2}\int_0^1\frac{\ln(1-\frac{\xi}{2})}{1-\frac{\xi}{2}}\text{ }d\xi\end{aligned}


    This second integral easily comes out to -\frac{1}{2}\ln^2(2). To do the first one merely note that


    \displaystyle \begin{aligned}-\int_0^1\frac{\ln(1-\frac{\xi}{2})}{\xi}\text{ }d\xi &= \int_0^1 \sum_{n=1}^{\infty}\frac{\xi^{n-1}}{2^n n}\\ &= \sum_{n=1}^{\infty}\frac{1}{2^n n^2}\end{aligned}


    But this is a fairly well-known series (EDIT: You actually did it in the second post!) and it sums to \frac{\pi^2}{12}-\frac{1}{2}\ln^2(2). Thus, adding together we get


    \displaystyle \sum_{n=1}^{\infty}\frac{H_n}{2^n n}=\frac{\pi^2}{12}



    \displaystyle\sum_{n=1}^\infty \frac{\text{H}_n}{n^2}
    Spoiler:


    We first note that that in our previous solution we derived that


    \displaystyle \frac{-\ln(1-z)}{1-z}=\sum_{n\in\mathbb{N}}H_n z^n


    so that


    \displaystyle \frac{1}{2}\ln^2(1-z)=\sum_{n\in\mathbb{N}}\frac{H_n}{n+1}z^{n+1}


    and thus


    \displaystyle \sum_{n\in\mathbb{N}}\frac{H_n}{(n+1)^2}=\frac{1}{  2}\int_0^1\frac{\ln^2(1-z)}{z}\text{ }dz


    Note though that letting x=\ln(1-z) transfers our integral into



    \displaystyle \int_0^{\infty}\frac{x^2}{1-e^{-x}}e^{-x}\text{ }dx=\int_0^{\infty}\frac{x^2}{e^x-1}\text{ }dx=2\zeta(3)


    where we've used the well-known identity \displaystyle \Gamma(s)\zeta(s)=\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\text{ }dx.


    Proof of identity:

    Spoiler:


    We recall that \displaystyle \Gamma(s)=\int_0^{\infty}t^{s-1}e^{-t}\text{ }dx. Thus, it's simple to check that


    \displaystyle \frac{\Gamma(s)}{n^s}}=\int_0^{\infty}e^{-n t}t^{s-1}\text{ }dt


    where the result follows by summing both sides over \mathbb{N} and recalling that \displaystyle \sum_{n\in\mathbb{N}}e^{-nt}=\frac{1}{e^t-1}



    Thus,


    \displaystyle \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2}=\zeta(3)


    Note lastly then that we have


    \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\frac{H_n}{n^2} &= 1+\sum_{n=2}^{\infty}\frac{H_n}{n^2}\\ &= 1+\sum_{n=2}^{\infty}\frac{H_{n-1}+\frac{1}{n}}{n^2}\\ &= \sum_{n=2}^{\infty}\frac{H_{n-1}}{n^2}+\zeta(3)\\ &= \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2}+\zeta(3)\\ &= 2\zeta(3)\\ &=2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2}\end{alig  ned}

    Remark: This last equality wasn't necessary just an astounding fact! I haven't thought about it too deeply, but I wonder if one could classify the set of f:\mathbb{N}\to\mathbb{N} such that \displaystyle \sum_{n\in\mathbb{N}}\frac{f(n)}{n^2}=2\sum_{n\in\  mathbb{N}}\frac{f(n)}{(n+1)^2} or any of the obvious generalizations.





    and one that involves much more algebra:

    \displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{3n+1}
    Spoiler:

    This is just simple algebra since


    \displaystyle \begin{aligned}\sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1} &=\sum_{n=0}^{\infty}\int_0^1 (-1)^n x^{3n}\text{ }dx\\ &=\int_0^1 \sum_{n=0}^{\infty}(-1)^n x^{3n}\text{ }dx\\ &=\int_0^1\frac{dx}{1+x^3}\\ &= \frac{1}{18}\left(2\sqrt{3}\pi+\ln(64)\right)\end{  aligned}


    Where the last part is gotten by factoring x^3+1=(x+1)(x^2-x+1), performing partial fractions, and then integrating the good old fashioned way.
    Last edited by Drexel28; December 26th 2010 at 02:52 PM.
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  11. #41
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    \displaystyle \begin{aligned}\sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1} &=\sum_{n=0}^{\infty}\int_0^1 (-1)^n x^{3n}\text{ }dx\\ &=\int_0^1 \sum_{n=0}^{\infty}(-1)^n x^{3n}\text{ }dx\\ &=\int_0^1\frac{dx}{1+x^3}\\ &= \frac{1}{18}\left(2\sqrt{3}\pi+\ln(64)\right)\end{  aligned}
    Nice work! From this, we can see in general for  a>0 ,  \displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{an+1} = \int_0^1 \frac{dx}{x^a+1} .
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  12. #42
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Nice work! From this, we can see in general for  a>0 ,  \displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{an+1} = \int_0^1 \frac{dx}{x^a+1} .
    Unfortunately those integrals are damn near impossible to compute even when a>5 and a\in\mathbb{N}! Since clearly the key is to see the cyclotomic factorziation and then use pfracs, but the non-linear term becomes intractable as you can clearly guess.
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  13. #43
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Here's my rebuttal:

    For  \alpha > 1 , evaluate  \displaystyle \lim_{t\to\infty} \frac t{\log t}\sum_{n=0}^\infty \frac1{\alpha^n+t} .
    Tough one!

    Let F(t) = \sum_{n=0}^\infty \frac{1}{\alpha^n+t}.

    We have, with t=\alpha^m, m>1, \frac t{\log t} F(t) = \frac{\alpha^m}{m\log \alpha} \sum_{n=0}^\infty \frac1{\alpha^n+\alpha^m} = \frac{1}{m \log \alpha}\sum_{n=0}^\infty \frac1{\alpha^{n-m}+1}.

    We can write this as \frac{1}{m \log \alpha}\left(\sum_{n=1}^m \frac1{\alpha^{-n}+1}+F(1)\right). Now F(1)/m \to 0 as m \to \infty hence to find the limit we can just look at \frac{1}{m \log \alpha} \sum_{n=1}^m \frac{\alpha^n}{1+\alpha^{n}}.

    Now we have \lim_{m\to \infty}\frac{1}{m}\sum_{n=1}^m \frac{\alpha^n}{1+\alpha^n} = 1 because \frac{1}{m}\sum_{n=1}^m \frac{1}{1+\alpha^n} < \frac{1}{m}\sum_{n=1}^m \frac{1}{\alpha^n} \to 0 as m \to \infty.

    Hence the desired limit is \frac{1}{\log \alpha}.
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  14. #44
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Tough one!

    Let F(t) = \sum_{n=0}^\infty \frac{1}{\alpha^n+t}.

    We have, with t=\alpha^m, m>1, \frac t{\log t} F(t) = \frac{\alpha^m}{m\log \alpha} \sum_{n=0}^\infty \frac1{\alpha^n+\alpha^m} = \frac{1}{m \log \alpha}\sum_{n=0}^\infty \frac1{\alpha^{n-m}+1}.

    We can write this as \frac{1}{m \log \alpha}\left(\sum_{n=1}^m \frac1{\alpha^{-n}+1}+F(1)\right). Now F(1)/m \to 0 as m \to \infty hence to find the limit we can just look at \frac{1}{m \log \alpha} \sum_{n=1}^m \frac{\alpha^n}{1+\alpha^{n}}.

    Now we have \lim_{m\to \infty}\frac{1}{m}\sum_{n=1}^m \frac{\alpha^n}{1+\alpha^n} = 1 because \frac{1}{m}\sum_{n=1}^m \frac{1}{1+\alpha^n} < \frac{1}{m}\sum_{n=1}^m \frac{1}{\alpha^n} \to 0 as m \to \infty.

    Hence the desired limit is \frac{1}{\log \alpha}.
    I did fairly the same thing except I proved that the limit existed. Namely, one can prove that if \displaystyle G(t)=\frac{t}{\log(t)}\sum_{n=0}^{\infty}\frac{1}{  \alpha^n+t} that for sufficiently large t one has G'(t)\leqslant 0. Moreover, one can show that \frac{1}{\log(\alpha)}\leqslant G(t) and thus it suffices to prove that G(\alpha^m)\overset{m\to\infty}{\longrightarrow}\f  rac{1}{\log(\alpha)}.
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  15. #45
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Have fun with this one (you won't, unless you 'see it')

    Problem 3: Compute

    \displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n \Gamma^3(x+1)\zeta^3(x+3)\text{ }dx
    How about a little hint on this one..
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