Is there a way to express this summation as an integral?

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The discussion centers on the feasibility of expressing the limit of a summation as an integral, specifically the equation \lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{R^2}{R^2+(4a\,i-2\,k)^2-(4a\,i-2\,k)\,\sin(\gamma)}. The participants conclude that the structure of the sum does not conform to the typical form that allows for convergence to an integral. The terms in the summation are independent of n, which prevents the necessary conditions for integral approximation from being met.

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gaganaut
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Hi,
This is to do with my research. While deriving some theory, I got an equation as follows.

[tex]\lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{R^2}{R^2+(4a\,i-2\,k)^2-(4a\,i-2\,k)\,\sin(\gamma)}[/tex]

Never mind what [tex]R, a, k,[/tex] and [tex]\gamma[/tex] are. They are all constants.

What I would like to do is to get a neat integral for this expression. So I would really appreciate if someone could let me know if doing this is even possible or not. The presence of the limit and the summation together tells me that there must be an integral solution.

I found a link where they deal with this stuff, but there, only a particular class of functions are dealt with (without saying so). Here is the URL,
http://johnmayhk.wordpress.com/2007/09/24/alpm-sum-an-infinite-series-by-definite-integrals/

Can something similar be worked out for this problem? If not is there any other general method to do this stuff. I don't expect a complete solution, but a small hint will work wonders.

Thanks.
 
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My first impression was that your equation does not look like an integral. The reason is that sums that converge to an integral generally look like

[tex]\sum_{i = 1}^{n} f(w_{i\,n})* \delta_{i\,n}[/tex]

where [tex]\delta_{i\,n} \rightarrow 0 \text{ as } n \rightarrow \infty[/tex]. Your sum does not look like that because the terms are independent of n. The first terms in your sum are always large, and then converge to 0 as i gets larger. This will allow the series to converge, but you will never get the first terms epsilon-small, which is what an integral does.
 
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