# Asymptotics of $\sum\limits_{j_1,\ldots, j_{2k}\neq n}\frac{1}{(n-j_1)(n-j_2)\cdots(n-j_{2k-1})(n-j_{2k})}$

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.

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What is that sum, again? The notation indicates all natural numbers $j$, $n$ with $j\ne n$, but then the sum is infinite. – Harald Hanche-Olsen Nov 16 '12 at 10:26
@HaraldHanche-Olsen the sum ranges over $j$ and I added in the statement that $n$ is fixed. Is it allright? – Ragnar Nov 16 '12 at 10:32
I think you really mean $j<n$. – user27126 Nov 16 '12 at 10:34
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. – Harald Hanche-Olsen Nov 16 '12 at 10:34
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? – Gerry Myerson Nov 16 '12 at 11:07

You donot say the lower bound of j and I suppose $j \ge 0$ and proved this: $\sum_{0\le j\le n}\frac{1}{|n-j|}=1+(H_n-1)\le 1+\int_1^n \frac{1}{x} dx=1+\log n$ and it is not difficult to prove the statement then.