I would like to calculate the following infinite series $a$.
$$a = \lim_{n\to\infty} \sum_{i=0}^{n-1} \frac{L - i \cdot \frac{2\cdot L}{n} }{\left[\left[ L-i\cdot \frac{2\cdot L}{n}\right]^2+y^2\right]^{3/2}}$$
Note, that I know also that $L$ is positive infinity large and much larger than $y\in\mathbb{R}_{>0}$.
I am not sure if my calculation is correct or not. Can somebody give me a hint if my way is not correct?
Here my way of solution
$$a = \lim_{n\to\infty} \sum_{i=0}^{n-1} \frac{L - i \cdot \frac{2\cdot L}{n} }{\left[L^2\cdot \left[ 1-i\cdot \frac{2}{n}\right]^2+y^2\right]^{3/2}}$$ $$= \lim_{n\to\infty} \sum_{i=0}^{n-1} \frac{L - i \cdot \frac{2\cdot L}{n} }{\left[L^2\cdot\left[\left[ 1-i\cdot \frac{2}{n}\right]^2+\frac{y^2}{L^2}\right]\right]^{3/2}}$$ $$= \lim_{n\to\infty} \sum_{i=0}^{n-1} \frac{L - i \cdot \frac{2\cdot L}{n} }{{L^2}^{3/2}\cdot\left[\left[ 1-i\cdot \frac{2}{n}\right]^2+\frac{y^2}{L^2}\right]^{3/2}}$$ $$= \lim_{n\to\infty} \sum_{i=0}^{n-1} \frac{L - i \cdot \frac{2\cdot L}{n} }{L^3\cdot\left[\left[ 1-i\cdot \frac{2}{n}\right]^2+\frac{y^2}{L^2}\right]^{3/2}}$$ $$= \lim_{n\to\infty} \sum_{i=0}^{n-1} \frac{L^3\cdot\left[\frac{1}{L^2} - i \cdot \frac{2}{n\cdot L^2} \right]}{L^3\cdot\left[\left[1-i\cdot \frac{2}{n}\right]^2+\frac{y^2}{L^2}\right]^{3/2}}$$ $$= \lim_{n\to\infty} \sum_{i=0}^{n-1} \frac{\frac{1}{L^2} - i \cdot \frac{2}{n\cdot L^2}}{\left[\left[1-i\cdot \frac{2}{n}\right]^2+\frac{y^2}{L^2}\right]^{3/2}}$$
Since, $L>>y$ and $L$ goes to infinity.
$$= \lim_{n\to\infty} \sum_{i=0}^{n-1} \frac{0 - 0}{\left[\left[1-i\cdot \frac{2}{n}\right]^2+0\right]^{3/2}} = \lim_{n\to\infty} \sum_{i=0}^{n-1} 0$$
So the limit while $n$ goes to infinity is
$$=0$$