I am trying to solve a summation formula that is quite complex. However, to make the "answering" process for you guys easier I'll isolate the part I am having trouble with...
The equation is as follows: $$ \lim_{h\to 0}\sum_{i=1}^{n/h} \frac{1}{\sqrt{ih(2c-ih)}} = \ ? $$ NOTE: 'c' is just a simple constant; if it helps just say 'c = 3' if you need it.
Let $\left[x\right]$ is the integer part of $x$. For partial summation you have $$\overset{n/h}{\underset{i=1}{\sum}}\frac{1}{\sqrt{ih\left(2c-ih\right)}}=\left[\frac{n}{h}\right]\frac{1}{\sqrt{n\left(2c-n\right)}}+\frac{1}{2}\int_{1}^{n/h}\left[t\right]\frac{h\left(2c-th\right)-th^{2}}{\left(th\left(2c-th\right)\right)^{3/2}}dt.$$ Note that $\left[t\right]\geq1;$ hence $$\overset{n/h}{\underset{i=1}{\sum}}\frac{1}{\sqrt{ih\left(2c-ih\right)}}\geq\left[\frac{n}{h}\right]\frac{1}{\sqrt{n\left(2c-n\right)}}+\frac{1}{2}\left(-\frac{1}{\sqrt{n\left(2c-n\right)}}+\frac{1}{\sqrt{h\left(2c-h\right)}}\right)\underset{h\rightarrow0}{\longrightarrow}+\infty.$$ P.S. I suppose we are working with real numbers, so I use a right-side limit.