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How to estimate the following sum in terms of $n$?

$$ \sum_{j_1,\ldots, j_{2k}\neq n}\frac{1}{(n-j_1)(n-j_2)\cdots(n-j_{2k-1})(n-j_{2k})}$$ with $n+j_1, j_1-j_2, \ldots, j_{2k}-j_{2k-1}, n-j_{2k} \in \{-2,2\}$.

Do you have any ideas and how to proceed?

Thanks.

  • 1
    What is that sum, again? The notation indicates all natural numbers $j$, $n$ with $j\ne n$, but then the sum is infinite.2012-11-16
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    @HaraldHanche-Olsen the sum ranges over $j$ and I added in the statement that $n$ is fixed. Is it allright?2012-11-16
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    I think you really mean $j.2012-11-16
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    But then the sum is infinite. But if @Sanches is correct, you just subsitute $j=n-k$ in your sum to get it on the form you already know.2012-11-16
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    To put it another way (and assuming $j$ is supposed to be at least zero), why don't you try writing out the sum explicitly for, say, $n=6$, just to see whether it looks familiar?2012-11-16
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    @Sanchez you are right, I edited it again. It's obvious that we obtain such a result with the substitution that you mentioned but can we improve $2 \log{n}$ ?2012-11-16
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    OP, isn't $\log n+\gamma+O(1/n)$ an improvement on $2\log n$?2012-11-17
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    @GerryMyerson if you don't take the modulus into account, yes it is.2012-11-17
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    You're summing over $j\lt n$, so $|n-j|$ is the same as $n-j$ ---- what does the modulus have to do with it?2012-11-17
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    I am now aware of the stupidty of this question. What I planned to ask was slightly different. I may post it as a new question. Thanks for your comments!2012-11-20
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    @беркай: Don't post a new question, improve this one instead.2012-12-04
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    I refined the question right now, here is my actual problem.2012-12-04
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    What sum do you mean? Suppose the inner sum is over $j_1$, then for all the other $j_i$ fixed the sum becomes a constant multiple of $$ \sum_{j_1\ne0}\frac1{j_1} $$ which either diverges, or converges to $0$ if we consider the Cauchy Principal Value $$ \lim_{N\to\infty}\sum_{|j_1|=1}^N\frac1{j_1} $$ So either the sum doesn't converge, or it converges to $0$.2012-12-04
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    Yes, it converges to $0$ but how fast? Can we express in terms of $n$?2012-12-04

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